- Tir n/du tifdng tfng: ,•^
n)ajna 3 (Bdl t^fi benzen tm ttSiiq, ttditq, ^Bai lq.f1 tdtig, huteêaebm
- ^Bai lq.f1 tdtig, huteêaebmi
t'Z:r-.:^^l BAI TAP MAU VA BAI TAP NANG CAÓ^.thn gsi^'^i ^=
i i .
Tim cong thiJc cau tao cua CgHs biet 3,12g chat nay phan iJng het vdi dung dich chu-a 4,8g Br2 hoac vdi toi da 2,688 lit H2 (dktc).
) Khi hidro hoa CgHg theo ti 1? mol 1 : 1 difdc hidrocacbon cilng loai X. Khi brom
h6a mot dong phan Y cua X vdi xuc t^c Fe (1 : 1) diTdc mpt sin pham duy nhat.
Xac dinh CTCT cua X , Ỵ
Ta cd : n _ 3 , 1 2 4,8 ii 111 t^lv Jiild^ J it = 0,03 mol mol H2: nH2 = = ^ = 0,12 mol , S 6 mol Br2: ngr, = — = 0,03 mol ^ 160 2.688 22,4 e l c d t i i ^ : ^ = M i 4 ; ^ = M 2 4 ' nê 0,03 1 nHj 0,12 4
Phan dgng phuong ph^p giai H6a hqc 1 1 HOu CO - B8 XuSn Hung
=> C T C T cua CsHs \k : [Q] stiren.
VI C R H X + H2 (1 : 1) X cdng loai => X ^ etylbenzen.
C H= C H 2 C H 2 - C H 3 '•'ô^r m,fc^'4r'')uMvi d:>ib -
^ ' ( X ) ' - ' ^
dong phan ciia X Ik Y .
Y + B r 2 (1 : 1) - » niQt san pham duy nhat => Y phai c6 C T C T dang doi xuTng.
n ; l C H 3
=> Y c6 C T C T m : (O) .(..ivn.vsYnị /KAN.,,-:!!./;;;- SC( no:, . - r . v
C H 3
B a i 2. Cho benzen tdc dung vdi lifdng dif HNO3 dSc c6 xiic tie H2SO4 dSc de dieu ché nitrobenzen. Tinh khoi lifdng nitrobenzen thu difdc khi dilng 1 tan benzen vdi hieu suat 78%. ' .
.H:>-'H Gidi : -^~5..H:,)..g;^
Phifdng trinh phan iJng : f r a n w f i m v
C H f i + HNO3 ) C 6 H 5 N O 2 + H 2 O
78 tan , ^ , „ . . 1 2 3 tan "5,i^S^~ .tjaSsSJj;
Khoi lifdng nitrobenzen : m = l i l i ? ! x — = 1,23 tan. IA ?j
• ^ 78 100
B a i 3 . . ... , . . . , - „;-