Giải chi tiết các câu liên quan đến anđêhit, xeton, axit cacboxylic trong đề thi đại học

Biên soạn GV Nguyễn Ngọc Sản Mobile: 01694043249 !"#$%&'()*+",+-. &/&01213&415)6 &17-8&2&9&01::/; ;/<=>& ?@ &1A)BC&.D&E&F&11G &.> )H/I) J)=H/:I) )K/I L)K/<I) S mol HCHO=s mol CO 2 =0,35 (mol) HCHOH 2 O 0,35 0,35 (mol) Suy ra s mol H 2 =(11,7:18)-0,35=0,3(mol) % th tớch H 2 =(0,3:0,65).100%=46,15% :/>%&E&M1@+"N>0O0OOP1' K &.' K &01= )QR&71/1+"N O0?,S1&71'/& A&9/:>+"8&N/> )3&1( 1T&"1"N1T&"11> ) U: ?A) J) UK ?A) ) ?A ?A)L) V: ?A) S mol Ag = 2 ln s mol anehit chng t anehit n chc S mol H 2 =2 ln s mol anehit chng t anehit cha no cha 1 pi ỏp ỏn A =:17113&415 W = W W ?M1@AW K = ?M1@/51"1A)J8& K = T>1-FX&Y)* 13&&71OZ01NOOP1' K &.' K &M.8&&G> )K) J)= ) L) CH=CH, HCHO, HCOOH, , 4 cht ny u tỏc dng vi AgNO 3 /NH 3 ỏp ỏn B =H,&11$,->1/M1T7)6 &17-&/K>/ &O01::/>& ?@&1A)'8&.[/K>&91CO\>OOP1':Q),&(> )/U U) J)/ K ) )/ L)/U) 0,3 mol axit trung ho 0,5 mol NaOH thỡ phi cú 1 axit n chc v mt axit hai chc n trung bỡnh l 5/3 nờn chn D ỏp ỏn D J =6 &17-&%&13&415/&01/K: /=KH<>& ?@&1A)J8&1( +"N?A &.T&.0]^()3&> ) K K ) J)_U_) ) _U U) L) ) -nCO 2 = nH 2 O = 0,197. X tỏc dng vi Cu(OH) 2, t 0 andehit no n chc ỏp ỏn D ,&251"1,&`1"1?2/`1(1\! -a&b11$A)& C$c)C%&&71OZ8&N'/!.=/=<>& ?@&1A)6 &17-&C/ !.H/= )T&"113&M&dC&.D^ >01G`&.>C>0&> )U U;/<;I) J)UH/I) )U U=/<<I) L)U=/<HI) - RCOOH + Na RCOONa + ẵ H 2 ; R(COOH) 2 + 2Na R(COONa) 2 + H 2 X x/2 y y nH 2 = x/2+y = 0,02(1) , nCO 2 = nx + ny = 0,06 (2) 1,5 <n < 3. Vy n = 2 (CH 3 COOH v HOOC-COOH) Thay vo (1), (2) x = 0,02 , y = 0,01 0,1.90 % .100 42,86(%) 0,1.90 0,2.60 HOOC COOH- = = + ỏp ỏn D :Le-17113&01!f,8&E1^&DOCg&%!T&h&.7!+> ) K / // K ) J) K // / K ) )/ K / / K L) K / // K ) t 0 sụi : CH 3 CHO < C 2 H 5 OH < HCOOH < CH 3 COOH. ỏp ỏn A Chuyên đề ôn thi đại học Hoá hữu cơ 2007-2013 Biên soạn GV Nguyễn Ngọc Sản Mobile: 01694043249 :17113&! ?A U ?$A U U ?1A U?AU ?OA K U?AU ?EA K U ?iA K UU K 7113&^&71OZ01N'/?A > )?1A/?OA/?iA J)?A/?$A/?1A )?A/?1A/?OA L)?1A/?OA/?EA Tỏc dng vi Na v Cu(OH) 2 Cú ớt nht 2 Nhúm OH cnh nhau (hoc COOH). ỏp ỏn C K13&415/51"1&71OZhGN:>OOP1/=Q/& 01%& KKH>5%&1>?@&1A)'8 &17-&>0&.a/!(3&Z8& !+Y17-$9#OOP1?A ?O0A&9 >0$9&DH/<)T&"11G 13&415&.> ) J) K K ) K L) K ; n X = n NaOH = 0,04; n ancol = 0,015< n X . 1 axit v 1 este ca axit ú. C n H 2 n O 2 n CO 2 + n H 2 O, 44x+18x =6,82. x= 0,11. n = nCO 2 /nX = 2,75 ; n.(0,04-0,015) +m.0,015 = 2,75.0,04 ; n= 2 (CH 3 COOH) v m = 4(CH 3 COOC 2 H 5 ) K:17113&415 ?:AW ?A1>/51"1/M1@W ?KA,1>W ?=AE&E/51"1/M1@W ?AEW ?HA1>T?1(%&>a8&T_A/M1@ ?;AW ?<AE&/51"1/M1@W ?A,&/51"1/M1@ ?:A,&T?1(%&>a8&T_A/51"1 Le-17113& &17-&^1! > $c! > > )?KA/?A/?HA/?<A/?A J)?KA/?=A/?HA/?;A/?:A )?A/?KA/?A/?;A/?A L)?:A/?KA/?A/?HA/?<A nCO 2 = NH 2 O CT chung hp cht C n H 2n O x Cú (3),(5),(6),(8),(9) tho món. ỏp ỏn A K13&4152>j8&8/^&71OZN'1(+"&.7$M1)J8&C &.D >0,&./2>C>0&>K/KKI=K/=I)T&"113&M1G2&05"> )U U UU U U U J)U? K AUU U )U UU U U L) K U K Th: loi B( khỏc dóy ), loi D(khụng tỏc dng vi Na). Cỏc cht trờn u cú CT chung C n H 2n O 2 32 53,33 32 43,24 % 2;% 3 14 32 100 14 32 100 X Y O n O m n m = = ị = = = ị = + + ỏp ỏn C K1>%&13&415?1"//A+"&N'R1N' K &9^ !.>)3&> )E&->E>1> J),&1 ),&KU.,.1 L)1>o-.,$Ek->1 Tỏc dng vi NaHCO 3 l axit: R(COOH) n + nNaHCO 3 + nCO 2 + Cú nX = nCO 2 1nhúm COOH. Tỏc dng vi Na nX = nH 2 cú thờm 1 nhúm OH. (axit 3-hidroxipropanoic: HO-CH 2 -CH 2 -COOH tho mn) A.HOCH 2 -CH 2 OH v D.o- HO-C 6 H 4 -CH 2 OH khụng td NaHCO 3 ; B.HOOC-(CH 2 ) 4 -COOH td vi NaHCO 3 -2a mol CO 2 . ỏp ỏn C 6 &17-&:>13&415/&01=> )3&&71OZ01N'/& +"&.7$M1+"1%J. &E&l>g>::)T&"113&M1G> )U_U J)U U U_U )U U U U L)U U_U nCO 2 = 4 cú 4C; T/dng vi Na cú -OH hoc -COOH; cú phn ng trỏng Ag cú -CH=O; cng Br 2 (1:1) cú 1liờn kt C=C CTCT l HO-CH 2 -CH=CH-CH=O Chuyên đề ôn thi đại học Hoá hữu cơ 2007-2013 Biªn so¹n GV NguyÔn Ngäc S¶n Mobile: 01694043249 Đáp án D : :Q1>/51"1,&11$,->151"12/^M1@1(1\! -a&b /&m! >1G13&>/>?! >1G2>N5! >1GA)'8 &17-&Q&9&01 KK/H>&  ?&1A/  )QR&71/8(QN  * = R1F&#1g+"E!&E7 ?g!3&><IA&9! E!&E&01> )K=/ J);/KH )/< L):</= Số C = n CO2 /n hh = 3 vậy ancol là C 3 H 7 OH → 4H 2 O . Vì n Nước < n CO2 nên axit không no. Axit có 3C có 2TH: CH 2 =CH-COOH → 2H 2 O ; x + y = 0,5 và 4x + 2y = 1,4. Ta có x= 0,2 và y = 0,3 (nhận) CH≡C-COOH → 1H 2 O ; x + y = 0,5 và 4x + y = 1,4. Ta có x= 0,3 và y = 0,2 (loại n Y < n X ) Este là CH 2 =CH-COOC 3 H 7 . Với m CH2=CH-COOC3H7 = 0,2*0,8*114 = 18,24 (g) ;,E&01^18$c171,71E],/!(&n&.OOP1  * = >e)6F &01:=,E&&9>01E1CO\?+!bg!3&7&.9^18M&;IA> )K J) )= L)H Sơ đồ : C 6 H 5 CH(CH 3 ) 2     O H O + + → (CH 3 ) 2 CO.      xeton cumen n mol n = = =      m g = = =</:>%&,&11$,->151"1/:> 1G,&(N>M^1(&m  >0>:/<)oa1G,&&.a> ),&.1) J),&E&1) ),&E&1) L),&$&1) Axit RCOOH, muối RCOOM: 0,1(R+45) + 0,1(R + M + 44) = 15,8 ⇒ 0,2R + 0,1M = 6,9. R = 1 ⇒ M = 67 (loại); R = 15 ⇒ M = 39 (K); R = 29 ⇒ M = 11(loại) . Vậy axit là CH 3 COOH (axit etanoic) ;1>E&->1,&11$,->1?/51"1/8&8&.Oe-jA &71OZ8&N'/+(.H/;>&  ?&1A)'8(?1(  * = R1>,S1&71A&9171 13&&.+"hGN&M&E!&E?+&8&+"E!&E(M&g !3&:IA),&&.> ) K ) J) K     ) )     K  ; ) L)   ;  =   ) N hh = 2n H2 = 0,6 (mol). Do axit tác dụng đủ với CH 3 OH nên n 2 axit = n CH3OH . = 0,3 (mol).    R COOH CH OH R COOCH H O − + → − + .  R COOCH M − = 25/0,3 = 83,33 ⇒ R = 24,33 (CH 3 - và C 2 H 5 -) Vậy 2 axit là CH 3 COOH và C 2 H 5 COOH. :J K !"#$#%&' ()*#+(+,-./% /0123(4526$"$"7(48#919:.  41!"  .%;5$#<#$#  $#" A. 0,015. =%% :%% >%% J+  -.   %  mol∑ = = = T1C!bOZW&1!+Y17- o1(,&&1/,&!&E.151"1a17-&M    . :.  ? 1[,&>>E1T1(>a8&T &. 151"1a17-1 ,& _   :.  .   − ⇒ ,&>>E1 _   @A@   ?   − = _/:> :BC$6#4DE!"F/  G/ F HI15#8#"%:$B2+J!K 4<!L++-.(48++E(5%/0123($B2+J!LM#8+8++N   $-   (48@N%:O!"P15#8< $B#" N%:    : !"P% B. C 2 H 3 COOH và 43,90%. :%:    : !"@P% >%: !"P% J+ BB&D+ >0 :>,&+"N'&9 >0&D ,::/U</      x mol − → = = Chuyªn ®Ò «n thi ®¹i häc Ho¸ h÷u c¬ 2007-2013  Biªn so¹n GV NguyÔn Ngäc S¶n Mobile: 01694043249 p9,&&0&.705a+1(?2A9Q 2 qQ  ) :  N  %@  ?  ? ?$#  % ⇒/:)=HV?/:U/:A?rV=A_</⇒r_;?   K A)ps-,&    K ?=K/IA K/4Q$4DERS!"4 TC$FHI5$##"5$#<UD <F%&526$"$"/(48V@#919:.  41!"  .%4 TC$F#" N%:  =%:    C. C 3 H 6 >%:    J+B&1!+Y17- a&?/51"1/M1@A    WR&71&1(       . :.  . :.  V@   ?$#W ? ?$#  ?    ⇒ ⇒2>ER1,1>) ?    E?≥AR1,1>?≥KAA QR&71  :. 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Thu vo 10,62 gam, mt i 18 gam , vy sau p khi lng dung dch gim = 18 10,62 = 7,38 gam. :$#6#oA4 $C_$]E!L4 (486##6# oA: : A: @ A: +Z#"(5] %&']E$"$"!L 6##6#*!K4<k#9+(+,g./%j2 ,<k#" )/;) =%% :%V@% >%% Hng dn: n axitsalixylic = 43,2/180 = 0,24 oUCH 3 COO-C 6 H 4 -COOH + 3NaOH CH 3 COONa + o-NaO-C 6 H 4 -COONa + 2H 2 O. 0,24 0,72 V NaOH = 0,72/1 = 0,72 lớt ;&526$"$"$#C$6#l(486$#:. !"_$# .!L_?6m% :$$#l2+J!L-:. +8(486$#:. %OQ<l#" N% 6#% J),&,>1) :%4%>%n$% Hng dn: S C trung bỡnh = nCO 2 /naxit = y/x S nhúm COOH = nCO 2 /naxit = y/x Cht cú s C = s nhúm chc ( loi A, C) Axit fomic chỏy cú nCO 2 = nH 2 O (loi) chn B :;&526$"$"C$6#ERS!"4a(HM #Q134T:?: $[\(48k#919:. 41!"6$# .%='(E#QoY2 2 ,6!"k#" N%k? x y % =%k? @ V x y )p_ x y+ ) >%k? @ V x y+ % Hng dn : Cụng thc chung ca cỏc axit trờn l : C n H 2n-4 O 4 C n H 2n-4 O 4 nCO 2 + (n-2)H 2 O T phng trỡnh ta thy : naxit = (nCO 2 nH 2 O)/2 naxit = (V/22,4 y)/2 Khi lng axit = xgam = mC/axit + mH/axit + mO/axit x = 12V/22,4 + 2y + 64(V/22,4 y)/2 x = 44V/22,4 30y V = 28/55(x + 30y) Chuyên đề ôn thi đại học Hoá hữu cơ 2007-2013 Biên soạn GV Nguyễn Ngọc Sản Mobile: 01694043249 Hn$!"$#%g$2+J!L-:. +87(48@#919:. 41%/0124526$"$"*V@#919. 41(48:. !"6$# .%j2 ,<6#" N%% =%% :%% L)/H) Hng dn: Cụng thc chung ca cỏc axit l R(COOH) X R(COOH) x + xNaHCO 3 R(COONa) x + xCO 2 + xH 2 O 0,7/x 0,7 n O/axit = 0,7.2 = 1,4 Axit + O 2 CO 2 + H 2 O 0,7 0,4 0,8 y Bo ton nguyờn t O : nO/axit + nO/O 2 = nO/CO 2 + nO/H 2 O 1,4 + 0,4.2 = 0,8.2 + y y = 0,6 <O ()C$6#$4DERSCp++-.T R$"CM++(]E(48(51%-3(4526$"$"7' 9$41*+Z#" N%#9% J)K/KH>&) :%#9% >%#9% Hng dn: n axit = (m mui m axit)/22 = (5,2 3,88)/22 = 0,06 M axit = 3,88/0,06 = 194/3 C n H 2n O 2 = 194/3 n = 7/3 C 7/3 H 14/3 O 2 + 5/2 O 2 7/3CO 2 + 7/3H 2 O. 0,06 0,15 V O 2 = 0,15.22,4 = 3,36 lớt. K&526$"$"4(48'919:. Cp'9D8L $Z4a( 1oo4M2(X%g$$#q+!LM#8+8++N $- 7(48 $#N%#" )E&i1) =%4$RSE% :%4% >%41T$RSE% Hng dn: Andehit t chỏy cú n CO 2 = n H 2 O andehit no, n nAg to ra / n andehit = 0,04/0,01 = 4 HCHO. =HHDM$4DE!"M$4EF5$# #LD5$#F(48M'9DCp'9<@- 4$Z $4a(1oo 4M2(X%-3(4526$"CM Q7(48#9:. 41%:O:O< F#*#8#"^ N%: A: A: !" :A: =%: A: !" :A: A: A: :%A: !" :A: L) K UU U Hng dn: nN 2 = nX = 5,6/28 = 0,2 nCO 2 = 0,48 C trung bỡnh = 0,48/0,2 = 2,4 (loi C) Dựng quy tc ng chộo da vo s C v C trung bỡnh v d kin s mol X ln hn s mol Y X l CH 3 COOH Da vo d kin s mol mi cht theo quy tc ng chộo v khi lng hn hp l 15,52 gam ta tỡm c Y l HOOC-CH 2 -COOH Chuyên đề ôn thi đại học Hoá hữu cơ 2007-2013 @ . cỏc axit trờn l : C n H 2n-4 O 4 C n H 2n-4 O 4 nCO 2 + (n-2)H 2 O T phng trỡnh ta thy : naxit = (nCO 2 nH 2 O)/2 naxit = (V/22,4 y)/2 Khi lng axit = xgam = mC /axit + mH /axit + mO /axit. ng trỏng Ag cú -CH=O; cng Br 2 (1:1) cú 1liờn kt C=C CTCT l HO-CH 2 -CH=CH-CH=O Chuyên đề ôn thi đại học Hoá hữu cơ 2007-2013 Biªn so¹n GV NguyÔn Ngäc S¶n Mobile: 01694043249 Đáp án D : :Q1>/51"1,&11$,->151"12/^M1@1(1!. D :Le-17113&01!f,8&E1^&DOCg&%!T&h&.7!+> ) K / // K ) J) K // / K ) )/ K / / K L) K / // K ) t 0 sụi : CH 3 CHO < C 2 H 5 OH < HCOOH < CH 3 COOH. ỏp ỏn A Chuyên đề ôn thi đại học Hoá hữu cơ 2007-2013 Biên soạn GV Nguyễn Ngọc Sản Mobile: 01694043249 :17113&! ?A U