Amines Test

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Amines Test - 55

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Weekly Quiz Competition

Question 1
1 / -0

Arrange the following in the decreasing order of the $\displaystyle{pK}_b$ values:
$i)\,\displaystyle { C}_{ 2 }{ H }_{ 5 }{ NH }_{ 2 }$
$ii) \,{ C }_{ 6 }{ H }_{ 5 }{ NHCH }_{ 3 }$
$iii) \,({ C }_{ 2}{ H }_{ 5 }{ ) }_{ 2 }NH$
$iv) { C }_{ 6 }{ H }_{ 5 }{ NH }_{ 2 }$

A
$i > ii > iii > iv$

B
$iii > i > ii > iv$

C
$i > iii > ii > iv$

D
$ii > i > iii > iv$

Solution

$+I$ effect of alkyl group increases electron density on $-N$ of amine and increases basic strength. Thus $(C_2H_5)_2NH$ is the strongest base. In aniline, the lone pair of electrons of $N$ is in delocalization with ring which makes it less available and least basic. Methyl substitution increases basic strength due to its inductive effect. Therefore the order of base strength or $pK_b$ is $iii>i>ii>iv$
Question 2
1 / -0

A compound $X$ has molecular formula ${C}_{7}{H}_{7}NO$ . On treatment with ${Br}_{2}$ and $KOH$ , $X$ gives an amine $Y$ . The later gives carbylamine test. $Y$ upon diazotisation and coupling with phenol gives an azo dye. Thus, $X$ is :

A
${C}_{6}{H}_{5}N{O}_{2}$

B
${C}_{6}{H}_{5}COON{H}_{4}$

C
${C}_{6}{H}_{5}CON{H}_{2}$

D
none of the above

Solution

1. $C_7H_7NO$ , show the high degree of unsaturation, so it can be considered that there may be phenyl group, as phenyl group has 3 double bond.

2. As the compound $C_7H_7NO$ , when reacts with $Br_2/KOH$ , gives the amine Y. It shows that there must be amide group in $C_7H_7NO$ compound, as this reaction is Hoffmann Bromamide synthesis.

3. As this amine is synthesized by Hoffmann Bromamide synthesis, it gives carbylamine test, so it means that amine must be primary amine.

4. Now as upon diazotisation this amine forms diazonium salt, also it gives the coupling reaction with phenol to give an azo dye, then it must be aromatic diazonium salt. If it were aliphatic then it would form alcohol by releasing $N_2$ , and the coupling could not be possible.

So, all these steps show that this is $C_6H_5CONH_2$ .
Reactions:
1. $C_6H_5CONH_2 (X) + Br_2 + 4 KOH\longrightarrow C_6H_5NH_2 (Y) + K_2CO_3 + 2 KBr + 2H_2O$

2. $C_6H_5NH_2 (Y) + CHCl_3 + 3KOH +\xrightarrow{heat}C_6H_5-NC + 3 KCl + 3H_2O$ .

3. $C_6H_5NH_2 (Y) \xrightarrow{NaNO_2 + 2HCl (at\ 273-278 K)}C_6H_5-N_2^{+}Cl^- + 2NaCl + 2H_2O$ .

4. $C_6H_5N_2^{+}Cl^- \xrightarrow {C_6H_5OH /OH-} C_6H_5-N=N-C_6H_5-OH (azo\ dye)$ .
Question 3
1 / -0

$CH_3CH_2CH_2CN+2H_2\xrightarrow {Pd}$

Which of the following compound is formed in the above reaction?

A
$CH_3CH_2CH_2CH_2NH_2$

B
$CH_3CH_2CH_2NH_2$

C
$CH_3CH_2CH_2CH_2CH_2NH_2$

D
None of the above

Solution

Alkyl nitriles are reduced using a metal like palladium, platinum or nicke as a catalyst.
So, $CH_3CH_2CH_2CN+2H_2\xrightarrow{Pd} CH_3CH_2CH_2CH_2NH_2$
Hence option $A$ is correct.
Question 4
1 / -0

In the reduction of nitriles to prepare primary amine:

A
$LiAlH_4$ is used.

B
$NaBH_4$ is used.

C
$Sn/HCl$ is used.

D
none of these.

Solution

$LiAl{ H }_{ 4 }$ reduces nitriles to primary amines with ascent of amine series.
${ Sn }/{ HCl }$ reduce nitriles to immine hydrochloride.
${ NaB{ H }_{ 4 } }/{ { I }_{ 2 } }$ reduces nitriles to primary amines
Question 5
1 / -0

Identify the compounds from the following which form primary amines under suitable reduction conditions.
1. $C_2H_5NC$
2. $C_2H_6$
3. $C_2H_5CONH_2$
4. $C_6H_5NO_2$

A
$1, 4$

B
$3, 4$

C
$1, 3, 4$

D
$2, 3, 4$

Solution

$C_2H_5CONH_2 \xrightarrow{LiAlH_4}C_2H_5CH_2NH_2$

$C_6H_5NO_2 \xrightarrow{Sn/HCl} C_6H_5NH_2$
Question 6
1 / -0

Predict the product.
$CH_3-CH_2-CN + H_2/Ni \rightarrow$ ?

A
$CH_3-CH_2-CH_2-NH_2$

B
$CH_3-CH_2-NH_2$

C
both a and b

D
none of these

Solution

$C{ H }_{ 3 }-C{ H }_{ 2 }-C\equiv N+{ { H }_{ 2 } }/{ Ni }\longrightarrow C{ H }_{ 3 }-C{ H }_{ 2 }-C{ H }_{ 2 }-N{ H }_{ 2 }$
Question 7
1 / -0

The reduction of nitrile gives ascent of amine series indicating:

A
preparation of amines containing two carbon atom more than starting amine

B
preparation of amines containing one carbon atom more than starting amine

C
preparation of amines containing three carbon atom more than starting amine

D
preparation of amines containing four carbon atom more than starting amine

Solution
Question 8
1 / -0

Predict respectively $'X'$ and $'Y'$ in the following reactions
$Ar-N{ H }_{ 2 }\xrightarrow { \quad \quad X\quad \quad } Ar-\overset { + }{ N } \equiv N.{ Cl }^- \xrightarrow { \quad \quad Y\quad \quad }$ $Ar-Cl$

A
$NaN{ O }_{ 3 }$ & ${ Cl }_{ 2 }$

B
$NaN{ O }_{ 3 }-HCl$ & $HCl$

C
$NaN{ O }_{ 2 }-HCl$ & ${ Cu }/{ HCl }$

D
$NaN{ O }_{ 2 }-HCl$ & $NaN{ H }_{ 2 }$

Solution

The balanced complete chemical equation is as follows:
$Ar-NH_2\xrightarrow{NaNO_2 - HCl} Ar-N^+_2-Cl^-$ , this reaction is the diazotisation reaction,
$Ar-N^+_2-Cl\xrightarrow{Cu/HCl} Ar-Cl + N_2$ , This is the Sandmeyer reaction.
Question 9
1 / -0

In the given set of reactions, the IUPAC name of product $Y$ is:

$2-\text{Bromopropane} \xrightarrow [ alc.\ KOH ]{ AgCN }\ \ X\xrightarrow [ ]{ LiAl{ H }_{ 4 } }\ \ Y\quad$

A
N-methylpropanamine

B
N-isopropylmethanamine

C
butan-2-amine

D
N-methylpropan-2-amine
Question 10
1 / -0

Point out the correct decreasing order of $pK_b$ values of following amines:
$C_2H_5NH_2$ , $C_6H_5NHCH_3,$ $(C_2H_5)_2NH$ and $C_6H_5NH_2$

A
$(C_2H_5)_2NH > C_2H_5NH_2 > C_6H_5NHCH_3 > C_6H_5NH_2$

B
$(C_2H_5)_2NH > C_6H_5NHCH_3 > C_6H_5NH_2 > C_2H_5NH_2$

C
$C_6H_5NH_2 > C_6H_5NHCH_3 > C_2H_5NH_2 > (C_2H_5)_2NH$

D
$C_2H_5NH_2 > (C_2H_5)_2NH > C_6H_5NHCH_3 > C_6H_5NH_2$

Solution

The lesser the value of $pK_b$ the weaker the base.
Electron withdrawing groups like benzene decreases the basicity and increases the $pK_b$ of amines and the alkyl groups (electron releasing groups) increases the basicity and decreases the $pK_b$ value
So the correct order of $pK_b$ is $C_6H_5NH_2 > C_6H_5NHCH_3 > C_2H_5-NH_2 > (C_2H_5)_2NH$
Hence option C is correct.

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Amines Test - 55

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Amines Test - 55

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Question 1

i>ii>iii>ivi > ii > iii > ivi>ii>iii>iv

iii>i>ii>iviii > i > ii > iviii>i>ii>iv

i>iii>ii>ivi > iii > ii > ivi>iii>ii>iv

ii>i>iii>ivii > i > iii > ivii>i>iii>iv

Solution

Question 2

C6H5NO2{C}_{6}{H}_{5}N{O}_{2}C6​H5​NO2​

C6H5COONH4{C}_{6}{H}_{5}COON{H}_{4}C6​H5​COONH4​

C6H5CONH2{C}_{6}{H}_{5}CON{H}_{2}C6​H5​CONH2​

none of the above

Solution

Question 3

CH3CH2CH2CH2NH2CH_3CH_2CH_2CH_2NH_2CH3​CH2​CH2​CH2​NH2​

CH3CH2CH2NH2CH_3CH_2CH_2NH_2CH3​CH2​CH2​NH2​

CH3CH2CH2CH2CH2NH2CH_3CH_2CH_2CH_2CH_2NH_2CH3​CH2​CH2​CH2​CH2​NH2​

None of the above

Solution

Question 4

LiAlH4LiAlH_4LiAlH4​ is used.

NaBH4NaBH_4NaBH4​ is used.

Sn/HClSn/HClSn/HCl is used.

none of these.

Solution

Question 5

1,41, 41,4

3,43, 43,4

1,3,41, 3, 41,3,4

2,3,42, 3, 42,3,4

Solution

Question 6

CH3−CH2−CH2−NH2CH_3-CH_2-CH_2-NH_2CH3​−CH2​−CH2​−NH2​

CH3−CH2−NH2CH_3-CH_2-NH_2CH3​−CH2​−NH2​

both a and b

none of these

Solution

Question 7

preparation of amines containing two carbon atom more than starting amine

preparation of amines containing one carbon atom more than starting amine

preparation of amines containing three carbon atom more than starting amine

preparation of amines containing four carbon atom more than starting amine

Solution

Question 8

NaNO3NaN{ O }_{ 3 }NaNO3​ & Cl2{ Cl }_{ 2 }Cl2​

NaNO3−HClNaN{ O }_{ 3 }-HClNaNO3​−HCl & HClHClHCl

NaNO2−HClNaN{ O }_{ 2 }-HClNaNO2​−HCl & Cu/HCl{ Cu }/{ HCl }Cu/HCl

NaNO2−HClNaN{ O }_{ 2 }-HClNaNO2​−HCl & NaNH2NaN{ H }_{ 2 }NaNH2​

Solution

Question 9

N-methylpropanamine

N-isopropylmethanamine

butan-2-amine

N-methylpropan-2-amine

Question 10

(C2H5)2NH>C2H5NH2>C6H5NHCH3>C6H5NH2(C_2H_5)_2NH > C_2H_5NH_2 > C_6H_5NHCH_3 > C_6H_5NH_2(C2​H5​)2​NH>C2​H5​NH2​>C6​H5​NHCH3​>C6​H5​NH2​

(C2H5)2NH>C6H5NHCH3>C6H5NH2>C2H5NH2(C_2H_5)_2NH > C_6H_5NHCH_3 > C_6H_5NH_2 > C_2H_5NH_2(C2​H5​)2​NH>C6​H5​NHCH3​>C6​H5​NH2​>C2​H5​NH2​

C6H5NH2>C6H5NHCH3>C2H5NH2>(C2H5)2NHC_6H_5NH_2 > C_6H_5NHCH_3 > C_2H_5NH_2 > (C_2H_5)_2NHC6​H5​NH2​>C6​H5​NHCH3​>C2​H5​NH2​>(C2​H5​)2​NH

C2H5NH2>(C2H5)2NH>C6H5NHCH3>C6H5NH2C_2H_5NH_2 > (C_2H_5)_2NH > C_6H_5NHCH_3 > C_6H_5NH_2 C2​H5​NH2​>(C2​H5​)2​NH>C6​H5​NHCH3​>C6​H5​NH2​

Solution

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$i > ii > iii > iv$

$iii > i > ii > iv$

$i > iii > ii > iv$

$ii > i > iii > iv$

${C}_{6}{H}_{5}N{O}_{2}$

${C}_{6}{H}_{5}COON{H}_{4}$

${C}_{6}{H}_{5}CON{H}_{2}$

$CH_3CH_2CH_2CH_2NH_2$

$CH_3CH_2CH_2NH_2$

$CH_3CH_2CH_2CH_2CH_2NH_2$

$LiAlH_4$ is used.

$NaBH_4$ is used.

$Sn/HCl$ is used.

$1, 4$

$3, 4$

$1, 3, 4$

$2, 3, 4$

$CH_3-CH_2-CH_2-NH_2$

$CH_3-CH_2-NH_2$

$NaN{ O }_{ 3 }$ & ${ Cl }_{ 2 }$

$NaN{ O }_{ 3 }-HCl$ & $HCl$

$NaN{ O }_{ 2 }-HCl$ & ${ Cu }/{ HCl }$

$NaN{ O }_{ 2 }-HCl$ & $NaN{ H }_{ 2 }$

$(C_2H_5)_2NH > C_2H_5NH_2 > C_6H_5NHCH_3 > C_6H_5NH_2$

$(C_2H_5)_2NH > C_6H_5NHCH_3 > C_6H_5NH_2 > C_2H_5NH_2$

$C_6H_5NH_2 > C_6H_5NHCH_3 > C_2H_5NH_2 > (C_2H_5)_2NH$

$C_2H_5NH_2 > (C_2H_5)_2NH > C_6H_5NHCH_3 > C_6H_5NH_2$