MCQ Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 5

What is the name for red azo dye?

0%
p-hydroxy azo benzene
0%
$\beta$ - nathyl azo benzene
0%
p-amino azo benzene
0%
p- N, N dimethyl amino azo benzene

Which of the following is the weakest Bronsted base?

0%
0%
0%
0%
$CH_3NH_2$

Explanation

Aniline is the weakest Bronsted base due to delocalisation of lone pair of electron of

$N-$ atom into benzene ring.

The correct order of basic strength of the following compounds is :

0%
2 > 1 > 3 > 4
0%
1 > 3 > 2 > 4
0%
3 > 1 > 2 > 4
0%
1 > 2 > 3 > 4

Explanation

Higher the electron density on electron donating group higher is the basic strength.

In compound

$3$ the electron density is highest because of

$+\ I$ effect of

$\text {methyl}$ groups attached to nitrogen. Hence it has the highest basicity.

In compound

$1$ there are two amine groups attached to the carbon in which one of the amine group is always available for donation. Hence it has the second highest basicity.

In compound

$2$ again the ethyl group increases the basic strength of amine due to its

$+\ I$ effect but lower than that of as in compound

$3$ .

The compound

$4$ has the least basic strength because the carbonyl group attached to amine decreases the basicity by withdrawing the lone pair from Nitrogen through resonance effect.

The correct order of basic strength is-

$3 > 1 > 2 > 4$

The correct order of increasing basicity of the given conjugate bases (R =

$CH_{3}$ ) is:

0%
$RCO\bar{O} < HC \equiv \bar{C} < \bar{R} < \bar{N}H_{2}$
0%
$\bar{R} < HC \equiv \bar{C} < RCO\bar{O} < \bar{N}H_{2}$
0%
$RCO\bar{O} < NH_{2} < HC \equiv \bar{C} < \bar{R}$
0%
$RCO\bar{O} < HC \equiv \bar{C} < \bar{N}H_{2} < \bar{R}$

Explanation

The order of acidity can be explained on the basis of the acids of the given conjugate base

$.$ Stronger is the acid

$,$ weaker is the conjugate base

$.$ Since

$,$

$RCOO$ is the stronger acid amongst all

$,$

$RCO{O}^{-}$ is the weakest base

$.$ Due to

$sp$ hybridised carbon

$,$ acetylene is also acidic and hence a weak base but stronger than

$RCO{O}^{-}.$ As

${sp}^{3}$ carbon is less electronegetive than

${sp}^{3}$ nitrogen

$,$

${R}^{-}$ is more basic than

$N{H}_{2}.$

What would be the product for the following reaction ?

0%
Ph-Cl
0%
$Ph-N_2$
0%
0%

Explanation

REF.Image.

$Na\, NO_{2} + HCl \rightarrow Nacl+ HNO_{2}$

$HO -N = O + H^{+} \Rightarrow H_{2}O^{\oplus }-N =0$

$H_{2}O^{\oplus}-N = 0 \rightarrow H_{2}O+N^{\oplus} =0$

1155283_877087_ans_5eb65e9b33b64dc58727dc6a66933f4d.png

The correct order of basicity of the following compounds is?

$\underset {(3)}{(CH_{3})_{4}NH}$

$\underset {(4)}{CH_{3}CONH_{2}}$

0%
$2 > 1 > 3 > 4$
0%
$1 > 3 > 2 > 4$
0%
$3 > 1 > 2 > 4$
0%
$1 > 2 > 3 > 4$

Explanation

$\begin{array}{l} \text { (4) lone pair is on resonance. } \\ \text { Hence, basisty order is } \\ \qquad 1>2>3>4 \end{array}$
2005258_877302_ans_3591d1dd86274f4287b1eab8748d8449.PNG

2005258_877302_ans_3591d1dd86274f4287b1eab8748d8449.PNG

Which of the following is not a primary amine?

0%
tert - Butylamine
0%
sec - Butylamine
0%
Isobutylamine
0%
Dimethylamine

Explanation

Dimethylamine

$[(CH_3)_2 NH]$ is a secondary amine as the central Nitrogen atom is substituted by two methyl groups and the nitrogen has one hydrogen. Tert-butyl amine, secondary butyl amine, isobutylamine are all the isomers of n-butyl amine, varying in the position of the

$N-$ group in the alkyl chain.

Secondary amines can be prepared by:

0%
reduction of nitro compounds
0%
reduction of amides
0%
reduction of isonitriles
0%
reduction of nitriles

Explanation

Isonitriles on reduction with lithium aluminium hydride

$(LiAlH_4)$ or catalytic hydrogenation (

$H_2/Ni$ ) produce secondary amines.

$R-NC \xrightarrow{LiAlH_4, H_2/Ni} R-NH-CH_2$

$C_3H_9N$ cannot represent:

0%
$1^o$ amine
0%
$2^o$ amine
0%
$3^o$ amine
0%
quaternary ammonium salt

Explanation

${C}_{3}{H}_{9}N$ can not represent a quaternary salt

$.$ Because it has only three carbon atom

$.$

Among the following which statement (s) is/are correct?

0%
Both N of pyrimidine are same basic strength.
0%
In imidazole protonation take places on N-3.
0%
In purine only one lone pair of N is delocalized.
0%
Pyrimidine, imidazole and purine all are aromatic.

Explanation

Pyrimidine has nitrogen atoms at 1 and 3 positions.

In purine only one lone pair of

$N$ is delocalized. But lone pair electron does not take place in delocalization and imidazole also.

Which one of the following is aromatic amine?

0%
0%
0%
0%
Both $A$ and $B$

Explanation

(A) Aniline is aromatic as it obeys Huckel's rule of aromaticity. It is aromatic.

(B) This is also followed Huckel's rule of aromaticity

$(4n+2)\pi e^-$

$=6\pi e^-(n=1)$

It is also planar, cyclic and conjugated system.

So both A and B are aromatic amines.

While option C is aliphatic amine since the

$NH_2$ group attached to aliphatic carbon.

Option D is correct.

Identify

$X,\ Y$ and

$Z$ in the given reaction.

$CH_2 \, =\, CH_2 \xrightarrow [CCl_4]{Br_2} \,X\, \xrightarrow[(2 \, moles)]{NaCN} \, Y \, \xrightarrow {LiAlH_4} \, Z$

0%
$X \,=\, CH_2BrCH_2Br \, ,\, Y \, =\, CH_3CH_2CH_2CN$
0%
$X \, =\, CH_2BrCH_2Br \, , \, Y \, =\, CH_3CH_2CN$
0%
$X\, =\, CH_3CH_2Br \, ,\, Y\, = \, CH_3CH_2CN$
0%
$X \, =\, CH_2BrCH_2Br \, ,\, Y \,=\, NCCH_2CH_2CN$

Explanation

$CH_2 = CH_2 \xrightarrow [CCl_4]{Br_2} \underset{(X)}{BrCH_2 - CH_2Br} \xrightarrow {NaCN} \underset {(Y)}{NCCH_2CH_2CN} \xrightarrow {LiAlH_4} \underset{(Z)} {\underset {(1,4-Diaminobutane)}{H_2NCH_2CH_2CH_2CH_2NH_2}}$

Tertiary amines have lowest boiling points amongst isomeric amines because:

0%
they have highest molecular mass
0%
they do not form hydrogen bonds
0%
they are more polar in nature
0%
they are most basic in nature

Which of the following explode on heating?

0%
Azides
0%
Sulphates
0%
Chlorides
0%
Phosphides

Gabriel phthalimide reaction is used for the preparation of __________ amines.

0%
primary aromatic
0%
secondary
0%
primary aliphatic
0%
tertiary

What is the correct order of basicity values of the following compounds?

0%
0%
$NH_3\,>\,CH_3NH_2\,>\,CH_2\,=\,CH-NH_2\,>\,NH_2-CH_2-CH_2-N^+R_3$
0%
0%
None

Explanation

$1.$

$NH_3\longrightarrow pk_a=34.5$

$2.$

$CH_3NH_4Br\longrightarrow pk_a=3.4$

$3.$

$CH_3NH_2\longrightarrow pk_a=10.6$

$4.$

$\begin{matrix} Me \\ Me \end{matrix}>\overset { + }{ S } - { CH }_{ 2 }-{ NH }_{ 2 } \rightarrow { pk }_{ a }=\sim 2$

Order is

$4>2>3>1$

Which of the following compounds will not undergo azo coupling reaction with benzene diazonium chloride?

0%
Aniline
0%
Phenol
0%
Anisole
0%
Nitrobenzene

Explanation

Diazonium cation is a weak electrophile and hence it reacts with electron-rich compounds containing electron-donating groups such as

$-OH$ ,

$-NH_2$ and

$OCH_3$ groups and not with compounds containing electron-withdrawing groups such as

$-NO_2,$ etc.

Hence, option D is correct.

The correct decreasing order of basic strength of the following species is:

$H_2O, \, NH_3, \, OH^-, \, NH_2^-$

0%
$NH_2^- \, > \, OH^- \,>\, NH_3 \, > \, H_2O$
0%
$OH^- \, >\, NH_2^- \,> \, H_2O \, > \, NH_3$
0%
$NH_3 \, > \, H_2O \, > \, NH_2^- \,> \, OH^-$
0%
$H_2O \, >\, NH_3 \, > \, OH^- \, > \, NH_2^-$

Explanation

$NH_3$ is more basic than

$H_2O$ , therefore

$NH^-_2$ (Conjugate base of weak acid

$NH_3$ ) is a stronger base than

$OH^-$ .Thus decreasing order of basic strength is

$NH^-_2$ >

$OH^-$ >

$NH_3$ >

$H_2O$ ,

In order to prepare a 1 amine from an alkyl halide with simultaneous addition of one

$CH_2$ group in the carbon chain, the reagent used as source of nitrogen is:

0%
Sodium amide $NaNH_2$
0%
Sodium azide $NaN_3$
0%
Potassium cyanide $KCN$
0%
Potassium phthalimide $C_6H_4(CO)_2N^-K^+$

Explanation

$4R-X \xrightarrow{KCN} R-CN \xrightarrow {Na\ / C_2H_5OH} R-CH_2NH_2$

Basic strength of different alkyl amines depends upon:

0%
$+I$ effect
0%
steric effect
0%
solvation effect
0%
all of these

Explanation

The basic strength of different alkyl amines depends on many factors like:

$+ I$ effect is the polarization of a sigma bond due to electron donating effect of adjacent groups or atoms.

2) In water, the ammonium salts of primary and secondary amines undergo solvation effects due to hydrogen bonding to a much greater degree than ammonium salts of tertiary amines.

3) steric effect is the effect faced by incoming group or hydrogen by the already present bulky

$-R$ groups on the Nitrogen atom.

Which of the following has highest

$pK_b$ value?

0%
$(CH_3)_3CNH_2$
0%
$NH_3$
0%
$(CH_3)_2NH$
0%
$CH_3NH_2$

Explanation

Higher the basicity, lower is the

$pK_b$ value.

Since

$NH_3$ is the weakest base, hence it has highest

$pK_b$ value.

Option B is correct.

Amongst the following, the strongest base in aqueous medium is ____________

0%
$CH_3NH_2$
0%
$NCCH_2NH_2$
0%
$(CH_3)_2NH$
0%
$C_6H_5NHCH_3$

Explanation

In aqueous medium ,

$2^0$ amine

$(CH_3)_2NH$ is more basic than

$1^0$ amine

$(CH_3)NH_2$ .Due to the -I effect of -CN group,

$NCCH_2NH_2$ is less basic than

$CH_3NH_2$ and due to the delocalisation of lone pair of electron of N atom in benzene ring.

$C_6H_5NHCH_3$ is less basic than

$CH_3NH_2$ and more basic than

$NC-CH_2NH_2$ .Thus, the strongest base in aqueous medium is

$(CH_3)_2NH$

Among the following amines, the strongest Bronsted base is:

0%
0%
$NH_3$
0%
0%

Explanation

Option

$D$ is the strongest Bronsted base as there is no delocalisation of lone pair of electron of

$N$ atom which is not possible in aniline and in pyrrole.

The correct order of boiling points of the following isomeric amines is:

$C_4H_9NH_2,(C_2H_5)_2NH,\, C_2H_5N(CH_3)_2$

0%
$C_2H_5N(CH_3)_2 \, >\, (C_2H_5)_2NH \, >\, C_4H_9NH_2$
0%
$(C_2H_5)_2NH\, >\, C_2H_5N(CH_3)_2 \,>\, C_4H_9NH_2$
0%
$C_4H_9NH_2 \, >\, (C_2H_5)_2NH \, >\, C_2H_5N(CH_3)_2$
0%
$(C_2H_5)_2 NH \, >\, C_4H_9NH_2 \, >\, C_2H_5N(CH_3)_2$

Explanation

Order of Boiling points of amines:

Primary amines (

$RNH_2$ ) > Secondary amines (

$R_2NH$ ) > Tertiary amines (

$R^{'}NR_2$ ).

Intermolecular hydrogen bonding is more in primary amines than that in secondary amines and nil in tertiary amines as there are no hydrogens attached to the nitrogen atom.

$C_4H_9NH_2 > (C_2H_5)_2NH > C_2H_5N(CH_3)_2$

The correct increasing order of basic strength for the following compounds is:

0%
II < III < I
0%
III < I < II
0%
III < II < I
0%
II < I < III

Explanation

Electron donating

$-CH_3$ group increases the basicity while electron withdrawing

$NO_2$ group decreases the basicity of amines.

Option D is correct.

Which (isomeric) amine has lowest boiling point?

0%
$1^o$ amine
0%
$2^$ amine
0%
$3^o$ amine
0%
cannot predict

Explanation

$3^o$ amine, there is no hydrogen directly attached to nitrogen, due to which hydrogen bonding is not possible in them. Due to this absence of hydrogen bonding, less energy is required for boiling leading to lower boiling point.

In sets a - d, only one of the set is incorrect regarding basic strength. Select it:

Explanation

The basic character of amines is because of the lone pair of nitrogen. More the availability of lone pair, more will be the basic strength.

In compound as shown in IMAGE

$01$ , the lone pair of nitrogen is involved in making the ring aromatic and hence is less available due to which this compound will be least basic.

The compound as shown in IMAGE

$02$ will be the most basic due to presence of

$2$ nitrogen i.e.,

$2$ lone pairs instead of

$1$ increase the basic strength.

The compound as shown in IMAGE

$03$ will be intermediate between theses compounds.

This order is not followed in option

$C$

1056220_989357_ans_9a9f087cc08b49b393bf9ed3e1ab3efb.PNG

Which of the following Diazonium salt is stable at room temperature?

Its basic strength is

$10^{10}$ more than 1-dimethyl amino naphthalene due to:

0%
resonance
0%
steric inhibitation of resonance
0%
ortho effect
0%
hyperconjugation

Explanation

For participation in resonance, the

$p-orbital$ of

$N$ should be perpendicular to the plane of the ring.
Presence of two bulky methyl groups does not allow the

$p-orbital$ to get perpendicular to the plane and hence the lone pair of

$N$ does not involve in resonance.
This increases the lone pair availability, thus increasing the basic strength.
Hence,

$1,8-bis(dimethylamino)$ napthalene is more basic than

$1-dimethyl amino$ napthalene due to steric inhibition of resonance.

Among the following compounds, the most basic compounds is:

Correct energy profile for amine inversion and hybridization of nitrogen in transition state is:

Explanation

Amine is

$sp^3$ hybridised and the transition state in

$Sp^2$ hybridised as the lone pair enters the nucleus leaving behind a vacant orbital. And after sometime, the lone pair popes out of the either side, since the inverted amine is an enantiomer, the

${\text {energy remains the same}}$ .
Option D is the correct curve.
1082173_988124_ans_69c782d57968412384695189dbf8a8f9.png

Diazotization of n-Bu-

$NH_2$ with

$NaNO_2/HCl$ gives _________ isomeric butene.

0%
$2$
0%
$3$
0%
$4$
0%
$5$

Among the isomeric amines select the one with the lowest boiling point.

Explanation

The boiling point of amine

$\propto$ tendency to form H-bonds.

No. of H-bonds -

$3^oAmine<2^oAmine<1^oAmine$ [In gas phase]

The

$1^o$ and

$2^o$ amine have less steric repulsion than

$3^o$ amine.

Here in

$3rd$ compound, it is a tertiary amine that is sterically hindered by the methyl group, that is the reason it has less tendency to form hydrogen bond and has a lowest boiling point.

The answer is C.

In the given pair identify most acidic compound in

$(A)$ and

$(B)$ . Most basic in

$(C)$ and

$(D)$ .

0%
$A - (I),\ B - (II),\ C - (I),\ D - (II)$
0%
$A - (II),\ B - (I),\ C - (I),\ D - (II)$
0%
$A - (II),\ B - (II),\ C - (II),\ D - (II)$
0%
$A - (I),\ B - (II),\ C - (I),\ D - (I)$

In the given pair of compounds, in which pair second compound has higher boiling point than first compound?

0%
0%
$HO - CH_2 - CH_2 - OH$ and $CH_3 - CH_2 - OH$
0%
0%

Correct order of basic strength of given amines is:

0%
$\underset{2^o}{Me_2NM} > \underset{1^o}{MeNH_2} > \underset{3^o}{Me_3N > NH_3}$ (Protic solvent)
0%
$\underset{2^o}{Et_2NH} > \underset{3^o}{Et_3H} > \underset{1^o}{EtNH_2} > NH_3$ (Protic solvent)
0%
$Me_3N > Me_2NH > Me - NH_2 > NH_3$ (Gas phase)
0%
All are correct

Explanation

Refer to Image

Basic strength (in gas phase)

Because Me is an electron donating group it increases the

$e^-$ density on N and hence increase its tendency to donate lone pair.

Basicity of amine is directly related to the stability of amine after gaining

$H^+$ .

In the aqueous phase, the substituted ammonium cations gets stabilized not only by e^- releasing effect of the alkyl group (+I) but also by selvation with water molecules. The greater the size of ions, lesser will be the selvation and he less stabilised is the ion. Secondly, when the alkyl group is small like

$-CH_3$ group, there is no steric hindrance to

$-H$ bounding.

$\therefore\quad \begin{matrix} Me_{ 2 }NH \\ 2^o \end{matrix}> \begin{matrix} Me_{ 2 }NH_2 \\1^o \end{matrix}> \begin{matrix} Me_{ 3 }N \\ 3^o \end{matrix}>NH_3$

$\therefore \quad \begin{matrix} Et_2NH \\ 2^o \end{matrix}>\begin{matrix} Et_3N \\ 3^o \end{matrix}>\begin{matrix} EtNH_2 \\ 1^o \end{matrix}>NH_3$

1082234_989387_ans_d0315791bfa24f0382e8b1faa02d3407.PNG

Which of the following cyclic amine has lowest

$\Delta G$ for inversion?

Explanation

More is the steric hindrance the N of amine, lesser is the

$\Delta G$ for inversion because it has the least tendency to invert)

Refer to Image

Most steric hindrance is in C. Therefore it has lowest

$\Delta G$ for inversion.

1082455_989425_ans_f311672c35dc44fba56b4e0b152d44ad.png

The amino ketone shown given undergoes a spontaneous cyclization on standing. What is the major product of this intramolecular reaction?

Which amine yield

$N-$ nitroso amine after treatment with nitrous acid

$(NaNO_2, HCl)?$

Explanation

$\bullet$ N-Nitroso amine is compound is which group is present.

$\bullet$

$R,{ R }^{ 1 }$ may be same or different alkyl or aryl group.

1119115_993170_ans_a4bc2b2f8d8941a6b1a59aa50f856858.png

Which of the following arylamines will not form a diazonium salt on reaction with sodium nitrite in hydrochloric acid?

0%
m-Ethylaniline
0%
p-Aminoacetophenone
0%
$4$ -Chloro- $2$ -nitroaniline
0%
N-Ethyl- $2$ -methyaniline

Explanation

$N-Ethyl-2methyl$ aniline is a

${ 2 }^{ \circ }$ -secondary aniline (aryl).

But the remaining are primary-aryl-amines which would result in dizonium salt formation.

1086378_993155_ans_fa9b5c38d5144855846f34861e2256a8.png

In which of the following reactions, backward reaction is favoured?

0%
$H - C \equiv C - H + Li^{+} {^-CH}_{2}CH_{3} \rightleftharpoons H - C \equiv C : ^{\ominus}Li^+ + H_3C - CH_3$
0%
0%
$CH_3CH_2\overset{+}{S}H_2 + CH_3CH_2OH \rightleftharpoons CH_3CH_2SH + CH_3CH_2\underset{H}{\underset{|}{\overset{\oplus}{H}}} - H$
0%

Explanation

$A$ ,

$H-C\equiv C-H$ is more acidic (due to

$sp$ character), hence forward reaction is favored.

$B$ ,

$2,2,2-trifluoroacetic$

$acid$ is more acidic than

$HOCH_2CH_3$ due to presence of electron-withdrawing fluorine atoms. Hence, forward reaction is favored.

$C$ ,

$CH_3CH_2\overset {+}{S}H_2$ losing a proton will be more favored because the proton can then attack on oxygen atom having lone pair in alcohol.

However in

$D$ ,

$Phenyloxonium$ losing a proton will be more favored as the proton can then attack on the lone pair of nitrogen in aniline. Also oxygen being more electronegative than nitrogen, positive charge on nitrogen is more favored than that on oxygen.

Hence, for

$D$ , backward reaction is favored.

Which of following is an example of Pinacol-Diazotization?

Explanation

Amine groups are converted to

$N_2^+$ (diazo) groups in presence of

$NaNO_2/HCl$ . When alcohols are adjacent to

$NH_2$ group Ketones/Aldehydes are formed.

For the diazonium ions the order of reactivity towards diazo-coupling with phenol in the presence of dilute

$NaOH$ is:

0%
I $<$ IV $<$ II $<$ III
0%
I $<$ III $<$ IV $<$ II
0%
III $<$ I $<$ II $<$ IV
0%
III $<$ I $<$ IV $<$ II

Explanation

The order of reacting towards diazo coupling with phenol in the presence of dilute

$NaOH$ of these diazonium ions is :-
1380568_996078_ans_fb0c345686544c2684460951af28500c.png

In which of the following compound the methylenic hydrogens are the most acidic?

0%
$CH_3COCH_2CH_3$
0%
$CH_3CH_2COOC_2H_5$
0%
$CH_3CH_2CH(COOC_2H_5)_2$
0%
$CH_3COCH_2CN$

Explanation

Here cyanide group pulled the bond pair electron from the chain by the inductive

$(-I$ effect

$)$ . So, hydrogen of methyl group are because more acidic in nature.

$C-H$ bond became labile due to pulling of electron by

$CN$ group.
1025855_992039_ans_97759e2cb75e46b4806de3b9cacfccc3.PNG

Which of the following diazonium salt is relatively stable of

$0-5^o$ C?

0%
$CH_3CH_2N\equiv N\}^{\oplus}Cl^-$
0%
$CH_3-C(CH_3)_2-N\equiv N\}^{\oplus} Cl^-$
0%
$CH_3-N\equiv N\}^{\oplus}Cl^-$
0%
$(CH_3)_3C-N\equiv N\}^{\oplus}Cl^-$

Which is the strongest base?

0%
Pyrole
0%
Aniline
0%
Pyridine
0%

Explanation

Hint: A molecule or ion which can easily share its lone pair of an electron are most basic

Correct Answer: Option

$D$

Explanation of Correct Option:

According to the lewis base concept, a stronger base can easily share or donates its lone pair of electrons.

In compound

$D$ , there is one lone pair of electrons present on

$N$ atom and it can be easily shared with other electrophilic ions or molecules.
Also, it has alkyl groups attached to it which donates by

$+I$ . hence it is the strongest base among all given compounds.

Explanation of incorrect Option:

In the case of compound $A$ . the lone pair of nitrogen involved in the aromaticity of the ring, hence cannot be shared to act as a base, therefore it does not act as a strong base.
. The nitrogen center of pyridine features a basic lone pair of electrons. This lone pair does not overlap with the aromatic $\pi$ system ring but directly attached with $sp{^2}$ hybrid carbon atoms, hence it is less basic than compound $D$ where the $N$ atom attached with $sp{^3}$ hybrid carbon atoms.
In the case of compound $B$ . the lone pair of nitrogen is involved in resonance with the ring so the lone pair on nitrogen will not be easily available for donation that is why aniline is a weak base.

Which of the following statement is not true?

0%
$CH_3CH_2NH_2$ is ethanamine
0%
$CH_2CH_2OH$ is an ether
0%
Cyclopentane and 2-pentene have a molecular formula of $C_5H_{10}$
0%
Alkenes an alkynes are unsaturated

Explanation

$C{H}_{3}C{H}_{2}N{H}_{2}$ is Ethanamine or Ethylamine

$C{H}_{2}C{H}_{2}OH$ is an alcohol and not an ester.

Cyclopentane and 2-pentene have Molecular formula

${C}_{5}{H}_{10}$

Alkenes and Alkynes are Unsaturated Compounds.

So, Option B is correct

What product is formed on heating two molecules of urea?

0%
Uric Acid
0%
Biuret
0%
Nitrogen
0%
Carbon dioxide

Explanation

Urea is

$NH_2-\overset{\overset{O}{||}}{C}-NH_2$

When

$2$ molecules of urea is heated, biuret is formed.

$2NH_2-\overset{\overset{O}{||}}{C}-NH_2 \longrightarrow NH_2-\overset{\overset{O}{||}}{C}-NH-\overset{\overset{O}{||}}{C}-NH_2$

Biuret

The most basic among the following is:

0%
${ H }_{ 3 }C-{ NH }_{ 2 }$
0%
${ CH }_{ 3 }-{ CH }_{ 2 }-NH_2$
0%
${ H }_{ 3 }C-NH-{ CH }_{ 3 }$
0%
${ H }_{ 3 }C-{ CH }(CH_3)-NH_2$

Explanation

${ H }_{ 3 }-C$

$\rightarrow -NH\leftarrow -C-{ H }_{ 3 }$

$+I$ effects of methyl group

Here, also methyl group attached with the N-atom. The inductive effect of methyl group increases the density of electron on N-atom. It becomes more basic.

Compare the boiling point of the following two amines.

0%
Boiling point of $I > II$
0%
Boiling point of $II > I$
0%
Both should have equal boiling points
0%
It can't be predicted

CBSE Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 5 - MCQExams.com

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Practice Class 12 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Medical Chemistry Organic Compounds Containing Nitrogen Quiz 5 - MCQExams.com

0:0:1

Practice Class 12 Medical Chemistry Quiz Questions and Answers

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