The IUPAC name of B is:
${(C{H_3})_2}CHCHO + C{H_3}MgBr\xrightarrow{{ether}}A\xrightarrow{{{H_3}{O^ + }}}B$
A.$2$-methylbutan-$3$-ol
B.Pentan-$2$-ol
C.$3$-methylbutan-$2$-ol
D.$2$-methylbutan-$2$-ol
Answer
279.6k+ views
Hint: Here, we need to complete this reaction in the presence of a Grignard reagent which is just an organo-metallic compound having halogen in it. It’s general formula is ${R^ - }M{g^{2 + }}{X^ - }$, where $R$ stands for any alkyl group or aryl group, $Mg$ is Magnesium and $X$ stands for halogen atom.
Complete answer:
The R group, that is, alkyl group acts as a strong base or as strong nucleophile, therefore, Grignard reagents can undergo nucleophilic addition reaction of acid-base reaction. It only reacts with those compounds in which acidic hydrogen is present or any carbonyl group is present.
Since in the above reaction, we have aldehyde so the reaction will take place with the carbonyl group present, that is, carbon oxygen double bond.
Let’s have a look at the basic mechanism of the Grignard reagent.
Step-$1$: Grignard reagent will form bond with carbonyl group:
$C{H_3}C{H_2}Mgr + RCOR \to C{H_3}C{H_2}C(OMgBr){R_2}$
Step-$2$: Addition of dilute acid which will lead to the hydrolysis:
$C{H_3}C{H_2}C(OMgBr){R_2} + {H_2}O\xrightarrow{{{H_3}{O^ + }}}C{H_3}C{H_2}C(OH){R_2} + Mg(OH)Br$
So, when aldehyde reacts with Grignard reagent then alcohol will be formed as a product.
And what kind of alcohol will be formed, that depends upon the precursor.
Now, with the help of the above given mechanism, let’s solve the above given reaction. And find out the IUPAC name of the compound B.
So,
${(C{H_3})_2}CHCHO + C{H_3}MgBr\xrightarrow{{ether}}{(C{H_3})_2}CHCH(OH)C{H_3}\xrightarrow{{{H_3}{O^ + }}}C{H_3}CH(C{H_3})CH(OH)C{H_3}$
So, the product A is ${(C{H_3})_2}CHCH(OH)C{H_3}$ and product B is $C{H_3}CH(C{H_3})CH(OH)C{H_3}$.
Therefore, the IUPAC name of B is Option-C, that is, $3$-methylbutan-$2$-ol.
Note:
Grignard reagents are known to be the organic compounds having high reactivity. It is necessary to keep them away from any moisture, as they can react with protons from any source to undergo acid-base reaction which is much faster as compared to the nucleophilic reaction.
Complete answer:
The R group, that is, alkyl group acts as a strong base or as strong nucleophile, therefore, Grignard reagents can undergo nucleophilic addition reaction of acid-base reaction. It only reacts with those compounds in which acidic hydrogen is present or any carbonyl group is present.
Since in the above reaction, we have aldehyde so the reaction will take place with the carbonyl group present, that is, carbon oxygen double bond.
Let’s have a look at the basic mechanism of the Grignard reagent.
Step-$1$: Grignard reagent will form bond with carbonyl group:
$C{H_3}C{H_2}Mgr + RCOR \to C{H_3}C{H_2}C(OMgBr){R_2}$
Step-$2$: Addition of dilute acid which will lead to the hydrolysis:
$C{H_3}C{H_2}C(OMgBr){R_2} + {H_2}O\xrightarrow{{{H_3}{O^ + }}}C{H_3}C{H_2}C(OH){R_2} + Mg(OH)Br$
So, when aldehyde reacts with Grignard reagent then alcohol will be formed as a product.
And what kind of alcohol will be formed, that depends upon the precursor.
Now, with the help of the above given mechanism, let’s solve the above given reaction. And find out the IUPAC name of the compound B.
So,
${(C{H_3})_2}CHCHO + C{H_3}MgBr\xrightarrow{{ether}}{(C{H_3})_2}CHCH(OH)C{H_3}\xrightarrow{{{H_3}{O^ + }}}C{H_3}CH(C{H_3})CH(OH)C{H_3}$
So, the product A is ${(C{H_3})_2}CHCH(OH)C{H_3}$ and product B is $C{H_3}CH(C{H_3})CH(OH)C{H_3}$.
Therefore, the IUPAC name of B is Option-C, that is, $3$-methylbutan-$2$-ol.
Note:
Grignard reagents are known to be the organic compounds having high reactivity. It is necessary to keep them away from any moisture, as they can react with protons from any source to undergo acid-base reaction which is much faster as compared to the nucleophilic reaction.
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