Tag: chemistry

Questions Related to chemistry

$CH _{3}COOH \xrightarrow[]{LiAlH _{4}}(A)\xrightarrow[H _{2}]{Ni} B$ In this reaction A and B respectively are :

  1. $CH _{3}OH and CH _{4}$

  2. $C _{2}H _{5}OH and C _{2}H _{6}$

  3. $CH _{3}CHO and C _{2}H _{5} OC _{2}H _{5}$

  4. $C _{2}H _{5}OH and CH _{3}OCH _{3}$


Correct Option: B
Explanation:

B
$LiAlH _{4}$ reduces ethanoic acid to ethanol and then $H _{2}$ / Ni reduces it further to alkane.

The reagent(s) used for converting ethanoic acid to ethanol is/are :

  1. $LiAlH _{4}$

  2. $BH _{3}$ in $THF$

  3. $PCl _{3}$

  4. $K _{2}Cr _{2}O _{7}/H^{+}$


Correct Option: A,B
Explanation:

$BH _3$ in $THF$ being a strong reducing agent, can efficiently convert carboxylic acid into the corresponding primary alcohol. Also, $LiAlH _4$ being a strong reducing agent, releases $H^-$ ions in enough quantity to convert carboxylic acid to the corresponding alcohol. So, ethanoic acid can be converted to ethanol using $BH _3$ in $THF$ and $LiAlH _4$.

Methyl alcohol when reacted with carbon monoxide using cobalt or rhodium as catalyst , compound $'A'$ is formed. $'A'$ on heating with HI in the presence of red phosphorus as catalyst $'B'$ is formed. $'B'$ is:

  1. $CH _3COOH$

  2. $CH _3CHO$

  3. $CH _3CH _2I$

  4. $CH _3CH _3$


Correct Option: D
Explanation:

$CH _3OH+CO\xrightarrow [ (High\quad Pressure) ]{ CO/Rh } \underbrace{ CH _3COOH} _{Compound A}$

Alcohol with carbon monoxide in presence of  $CO/Rh$ gives carboxylic Acid.
$CH _3COOH\quad \underrightarrow{Red P+HI}\quad CH _3-CH _3$
$Red P+HI$  is areducing agent and reduces carboxylic acid into hydrocarbons.

Alcohols can be obtained from carboxylic acids by reduction using one of the following reagents:

  1. $LiAIH _{4}$

  2. $N/alcohol$

  3. $Zn/HCl$

  4. $H _{2}/Ni$


Correct Option: A
Explanation:

$RCOOH  +  LiAlH _4 \rightarrow RCHO  \rightarrow RCH _2OH $

Carboxylic acids along with lithium aluminium hydride, initially gives aldehyde and then reduced to primary alcohols.

Which optically active compound on reduction with $\displaystyle LiAlH _{4}$ will give optically inactive compound?

  1. $CH _3-CH(OCH _3)-COOH$

  2. $CH _3-CH _2-CH(OH)-COOH$

  3. $CH _3-CH _2-CH(CH _2OH)-COOH$

  4. $CH _3-CH(OH)-CH _2-COOH$


Correct Option: C

The organic product formed in the reaction. $C _{6}H _{5}COOH \xrightarrow[(II)H _{3}O]{(I)LiAIH _{4}}$?

  1. $C _{6}H _{5}CH _{2}OH$

  2. $C _{6}H _{5}OHCH _{4}$

  3. $C _{6}H _{5}CH _{3} and CH _{3} OH$

  4. $C _{6}H _{5}CH _{3} and CH _{4} $


Correct Option: A
Explanation:

using LAH (lithium aluminium hydride), the COOH group is reduced to  $ CH _{2} OH$  group

What happens when a carboxylic acid is treated with lithium aluminium hydride?

  1. Aldehyde is formed.

  2. Primary alcohol is formed.

  3. Ketone is formed.

  4. Grignard reagent is formed.


Correct Option: B
Explanation:

$RCOOH + 4H \xrightarrow{LiAlH _4}RCH _2OH+H _2O$

Consider the following reaction $CH _3-(CH _2) _{14}-COOH \overset{LiAlH _4}{\longrightarrow} \,X \overset{HCl}{\longrightarrow}$
Palmitic acid
$ Y \overset{1.\, Mg/Et _2O}{\underset{2.\, Oxirane}{\longrightarrow}} Z \overset{KMnO _4}{\longrightarrow} W$
The correct identity of the compounds is/are :

  1. $X \,is \,CH _3-(CH _2) _{14}-CH _2OH$

  2. $Y\, is \, CH _3-(CH _2) _{14}-CH _2Cl$

  3. $Z \,is\, CH _3-(CH _2) _{16}-CH _2OH$

  4. $W \, is\, CH _3-(CH _2) _{16}-CHO$


Correct Option: A,B,C
Explanation:

Reaction:
$CH _3-(CH _2) _{14}-COOH \overset{LiAlH _4}{\longrightarrow} \,CH _3-(CH _2) _{14}-CH _2OH \overset{HCl}{\longrightarrow} CH _3-(CH _2) _{14}-CH _2Cl$
$ CH _3-(CH _2) _{14}-CH _2Cl \overset{1.\, Mg/Et _2O}{\underset{2.\, Oxirane}{\longrightarrow}} CH _3-(CH _2) _{16}-CH _2OH \overset{KMnO _4}{\longrightarrow} CH _3-(CH _2) _{16}-COOH$

The order of reactivities of methyl halides in the formation of Grignard reagent is :

  1. $CH _3I> CH _3Br> CH _3Cl$

  2. $CH _3Cl> CH _3Br> CH _3I$

  3. $CH _3Br> CH _3Cl> CH _3I$

  4. $CH _3Br> CH _3I> CH _3Cl$


Correct Option: A
Explanation:

Grignard reagents are highly reactive and react with any source of a proton to give hydrocarbons. Water, alcohol, amines are sufficiently acidic to convert them to corresponding hydrocarbons. Halide reactivity increases in the order $Cl < Br < I$

Therefore A is the correct option. 

Grignard reagent, a very useful starting compound for a number of organic reaction can be prepared by:

  1. reaction of alkyl halides with a solution of magnesium hydroxide

  2. reaction of alkyl halide with dry magnesium powder in presence of dry ether

  3. reaction of $MgCl _2$ with ether and alcohol

  4. rection of alkyl halide with magnesium in presence of alcohol


Correct Option: B
Explanation:

$R-X+Mg\xrightarrow{dryether} R-Mg-X$
                                        grignard reagent