Organic Compounds Containing Halogens Question 165
Question: The major product formed in the following reaction is $ \underset{H}{\overset{CH_3}{\mathop{CH_3-\underset{|}{\overset{|}{\mathop{C}}}-CH_2}}}\ Br\ \ \underset{CH_3OH}{\mathop{\xrightarrow{CH_3O}}} $
[AIIMS 2005]
Options:
A) $ \underset{H}{\overset{CH_3}{\mathop{CH_3-\underset{|}{\overset{|}{\mathop{C}}}-CH_2}}}OCH_3 $
B) $ CH_3\underset{OCH_3}{\mathop{-\underset{|}{\mathop{C}}H-}}CH_2\ CH_3 $
C) $ \overset{CH_3}{\mathop{CH_3-\overset{|}{\mathop{C}}=CH_2}} $
D) $ \underset{OCH_3}{\overset{CH_3}{\mathop{CH_3-\underset{|}{\overset{|}{\mathop{C}}}-CH_3}}} $
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Answer:
Correct Answer: D
Solution:
$ \underset{\underset{H}{\mathop{|}}}{\mathop{\overset{\overset{CH_3}{\mathop{|}}}{\mathop{H_3C-C-CH_2-Br}}}}\underset{CH_3OH}{\mathop{\xrightarrow{CH_3{O^{-}}}}}A $ -
Alkyl halide is 1°. Keep in mind 1° halide gives product by SN 2 / E 2 mechanism and 1° halide always gives substitution reaction except when strongly hindered base is used. ex.: With $ \underset{\underset{CH_3}{\mathop{|}}}{\mathop{\overset{\overset{CH_3}{\mathop{|}}}{\mathop{CH_3-C-{O^{(-)}}}}}} $ it gives mainly elimination.
The reaction involves carbocation intermediate. i.e. $ \underset{\text{(primary carbocation)}}{\mathop{\underset{\underset{H}{\mathop{|}}}{\mathop{\overset{\overset{CH_3}{\mathop{|}}}{\mathop{CH_3-C-C{{\overset{\oplus }{\mathop{H}}}_2}}}}}}} $ but as it is a primary carbocation it will rearrange to give a tertiary carbocation, which completes the reaction $ \underset{\text{tertiary carbocation}}{\mathop{\underset{\underset{CH_3}{\mathop{|}}}{\mathop{\overset{\overset{CH_3}{\mathop{|}}}{\mathop{CH_3-{C^{\oplus }}}}}}}} $ Stability of carbocation: $ 3{}^\circ >2{}^\circ >1{}^\circ > $
$ \overset{\oplus }{\mathop{C}}H_3 $
It is because the stability of a charged system is increased by dispersal of the charge. The more stable the carbocation, the faster it is formed. N.B. - Rearrangement can occur via two mechanisms. Therefore,
BETA
