Single Choice

Choose the correct answer among the alternatives given: In the following reactions, (i)$$CH_3-\overset{CH_3}{\overset{|}C}H-\underset{OH}{\underset{|}C}H-CH_3\xrightarrow{H^+/heat}\underset{[Major\ product]}A +\underset{[Minor\ product]} B$$ (ii)$$A\xrightarrow[in\,absence\,of \,peroxide]{HBr,dark}\underset{Major\ product}C+\underset{Minor\ product}D$$ The major products $$A$$ and $$C$$ are respectively?

A$$CH_2=\overset{CH_3}{\overset{|}C}-CH_2-CH_3$$ and $$\underset{Br}{\underset{|}C}H_2-\overset{CH_3}{\overset{|}C}H-CH_2-CH_3$$
B$$CH_3-\underset{CH_3}{\underset{|}C}=CH-CH_3$$ and $$CH_{ 3 }-\overset { \overset { Br }{ | } }{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } -CH_{ 2 }-CH_{ 3 }$$
Correct Answer
C$$CH_3-\overset{CH_3}{\overset{|}C}=CH-CH_3 $$ and $$CH_3-\overset{CH_3}{\overset{|}C}H-\underset{Br}{\underset{|}C}H-CH_3 $$
D$$CH_2=\overset{CH_3}{\overset{|}C}-CH_2-CH_3$$ and $$CH_3-\overset{CH_3}{\overset{|}{\underset{Br}{\underset{|}C}}}-CH_2-CH_3$$

Solution

(i)$$CH_3-\overset{CH_3}{\overset{|}C}H-\underset{OH}{\underset{|}C}H-CH_3\xrightarrow{H^+/heat}\underset{[Majorproduct]}A +\underset{[Minorproduct]} B$$

its dehydration reaction of alcohol that gives mixture of alkene where the major product is the more substituted alkene according to Saytzeff rule and minor product is the less substituted alkene as:

$$CH_3-\overset{CH_3}{\overset{|}C}H-\underset{OH}{\underset{|}C}H-CH_3\xrightarrow{H^+/heat}\underset{[Major\ product=A]}{CH_{ 3 }-\underset { CH_{ 3 } }{ \underset { | }{ C } } =CH-CH_{ 3 } }+\underset{[Minor\ product=B]} {CH_{ 3 }-\underset { CH_{ 3 } }{ \underset { | }{ C } } HCH=CH_{ 2 }}$$

(ii)$$A\xrightarrow[in\,absence\,of \,peroxide]{HBr,dark}\underset{Majorproduct}C+\underset{Minorproduct}D$$

Its a addition reaction that follows Markovnikov's rule and H goes to the double bonded C having more number of hydrogen. Thus
$$CH_{ 3 }-\underset { CH_{ 3 } }{ \underset { | }{ C } } =CH-CH_{ 3 } +HBr \rightarrow \underset{[Major\ product=C]}{CH_{ 3 }-\overset { \overset { Br }{ | } }{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } -CH_{ 2 }-CH_{ 3 }}+\underset{[Minor\ product=D]}{CH_{ 3 }-\overset { \overset { H }{ | } }{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } -\overset { \overset { Br }{ | } }{ C } H-CH_{ 3 }}$$
Therefore A and C are $$CH_{ 3 }-\underset { CH_{ 3 } }{ \underset { | }{ C } } =CH-CH_{ 3 } $$ and $$CH_{ 3 }-\overset { \overset { Br }{ | } }{ \underset { \underset { C{ H }_{ 3 } }{ | } }{ C } } -CH_{ 2 }-CH_{ 3 }$$

SIMILAR QUESTIONS

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In the following dehydration of diol with $$\displaystyle { H }_{ 3 }{ PO }_{ 4 }$$, following product is:

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