Some Basic Principles Of Organic Chemistry Question 6
Question: A straight chain hydrocarbon has the molecular formula $ C_8H_{10} $ . The hybridisation for the carbon atoms from one end of the chain to the other are respectively $ sp^{3},sp^{2},sp^{2}\text{, }sp^{3},sp^{2},sp^{2},sp $ and $ sp $ . The structural formula of the hydrocarbon would be [CBSE PMT 1992]
Options:
A) $ CH_3-C\equiv C-CH_2-CH=CH-CH=CH_2 $
B) $ CH_3-CH_2-CH=CH-CH_2-C\equiv C-CH=CH_2 $
C) $ CH_3-CH=CH-CH_2-C\equiv C-CH=CH_2 $
D) $ CH_3-CH=CH-CH_2-CH=CH-C\equiv CH $
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Answer:
Correct Answer: D
Solution:
$ \underset{sp^{3}}{\mathop{CH_3}},-\overset{sp^{2}}{\mathop{CH}},=\underset{sp^{2}}{\mathop{CH}},-\overset{sp^{3}}{\mathop{CH_2}},-\underset{sp^{2}}{\mathop{CH}},=\overset{sp^{2}}{\mathop{CH}},-\underset{sp}{\mathop{C}},\equiv \underset{sp}{\mathop{CH}}, $
