Question

Isobutyl magnesium bromide with dry ether and ethyl alcohol gives:

• A. $CH_{3} -\underset{\overset{|}{CH_{3}}}{CH}-CH_{2} OH$and $CH_{3} CH_{2} MgBr$
• B. $CH_{3} -\underset{\overset{|}{CH_{3}}}{CH}-CH_{3} $and $MgBr(OC_{2}H_{5})$
• C. $CH_{3} -\underset{\overset{|}{CH_{3}}}{CH}-CH=CH_{2}$and $ Mg (OH)Br$
• D. $CH_{3} -\underset{\overset{|}{CH_{3}}}{CH}-CH_{3}$and $CH_{3}CH_{2} OMgBr$

Answer 1

Answer: (B)

$(CH_{3})_{2} CHCH_{2}MgBr \xrightarrow{C_{2}H_{5}OH} (CH_{3})_{2} CHCH_{3} +C_{2} H_{5} OMgBr$
as $R$ of Grignard’s reagent will take proton and form alkane.