MCQ
When $CH_3-CH_2-COOH$ is reduced with $LiAlH_4$, the compound obtained will be
- A$CH_3CH_2COOH$
- B$CH_3CH_2CHO$
- ✓$CH_3CH_2CH_2OH$
- D$H_2C=CH-CH_2-OH$
✓
Answer
Correct option: C.
$CH_3CH_2CH_2OH$
c
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
$MnO_4^{2 - }$ ($1$ mole) in neutral aqueous medium is disproportionate to
→Which of the following is not an element
Identify major product ‘P’ formed in the following reaction.
End product of the reaction is
Which of the following is known as freon which is used as a refrigerant
Spin magnetic moment of a octahedral complex having $CFSE = -0.8\Delta_0$ and surrounded by weak field ligands can be
→Total number of unpaired electrons in an atom of atomic number $29 $ is
→Assertion : Equal moles of different substances contain same number of constituent particles.
Reason : Equal weights of different substances contain the same number of constituent particles.
Reason : Equal weights of different substances contain the same number of constituent particles.
Which of the following species would be expected paramagnetic
The surface of copper gets tarnished by the formation of copper oxide. $N _2$ gas was passed to prevent the oxide formation during heating of copper at $1250 K$. However, the $N _2$ gas contains $1$ mole $\%$ of water vapour as impurity. The water vapour oxidises copper as per the reaction given below:$2 Cu ( s )+ H _2 O ( g ) \longrightarrow Cu _2 O ( s )+ H _2( g )$
$P _{ H _2}$ is the minimum partial pressure of $H _2$ (in bar) needed to prevent the oxidation at $1250 K$.
The value of $\ln \left( p _{ H _2}\right)$ is $. . . . .$
(Given: total pressure $=1$ bar, $R ($universal gas constant$) =8 JK ^{-1} mol ^{-1}, \ln (10)=2.3$.
$Cu ( s )$ and $Cu _2 O ( s )$ are mutually immiscible.
At $1250 K : 2 Cu ( s )+1 / 2 O _2( g ) \longrightarrow Cu _2 O ( s ) ; \Delta G ^\theta=-78,000 J mol ^{-1}$
$H _2( g )+1 / 2 O _2( g ) \longrightarrow H _2 O ( g ) ; \Delta G ^\theta=-1,78,000 J mol ^{-1} ; G$ is the Gibbs energy$)$
→$P _{ H _2}$ is the minimum partial pressure of $H _2$ (in bar) needed to prevent the oxidation at $1250 K$.
The value of $\ln \left( p _{ H _2}\right)$ is $. . . . .$
(Given: total pressure $=1$ bar, $R ($universal gas constant$) =8 JK ^{-1} mol ^{-1}, \ln (10)=2.3$.
$Cu ( s )$ and $Cu _2 O ( s )$ are mutually immiscible.
At $1250 K : 2 Cu ( s )+1 / 2 O _2( g ) \longrightarrow Cu _2 O ( s ) ; \Delta G ^\theta=-78,000 J mol ^{-1}$
$H _2( g )+1 / 2 O _2( g ) \longrightarrow H _2 O ( g ) ; \Delta G ^\theta=-1,78,000 J mol ^{-1} ; G$ is the Gibbs energy$)$