MCQ Questions for Class 11 Medical Chemistry Organic Chemistry Some Basic Principles And Techniques Quiz 10

Which of the following has

$p_π−d_π$ bonding?

0%
$NO_3^-$
0%
$SO_3^{2-}$
0%
$BO_3^{3-}$
0%
$CO_3^{2-}$

Number of bonds in

$SO_2$ :

0%
Two $\sigma$ and two $\pi$
0%
Two $\sigma$ and one $\pi$
0%
Two $\sigma$ and two $\pi$ and one lone pair
0%
None of these

Explanation

Answer is Option C.

The sulphur dioxide molecule contains

$2σ$ ,

$2π$ bonds and one lone pair of electrons.

$S$ atom contains

$6$ valence electrons out of which it shares

$2$ electrons with

$2$

$O$ atoms to form

$2$

$sigma$ bonds.It also shares

$2$ electrons with

$2$

$O$ atoms to form

$π$ bonds. The remaining

$2$ electrons appear as

$one$ lone pair of electrons. $$SO_2$$ molecule has bent geometry. The

$S$ atom undergoes $$sp^2$$ hybridization.

1671718_1752268_ans_9ed0c1ed58644e588dcce5e61e32fdc0.jpg

The number of

$\pi$ bonds in 3-hexyne-1-ene is

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

Explanation

$\overset{1}{CH_2}\overset{1\ \pi}=\overset{2}{CH}-\overset{3}{C}\overset{2\ \pi}{\equiv} \overset{4}{C}-\overset{5}{CH_2}-\overset{6}{CH_3}$

3-hexyne-1-ene

The number of

$\pi$ bonds in 3-hexyne-1-ene is 3.

Which of the following is not o, pdirecting group

0%
$-NH_{2}$
0%
$-OH$
0%
$-X (halogens)$
0%
$-CHO$

Explanation

$\text{The presence of}$

$-CHO$

$\text{decreases electron density at o to p positions. Hence, the attack of an}$

$\text{electrophile occurs at m -position. therefore it is meta directing group.}$
1624732_1764812_ans_be65d862e0aa4a479e8f40da83403531.jpg

How many

$\pi$ bonds are present in naphthalene molecule?

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

$5\pi$ bonds are present in naphthalene.
1678366_1761305_ans_bf64497e341642c78cef8bca713b5b70.png

Which of the following is azo-group

0%
$-N=$
0%
$-N=N-$
0%
$-NH-$
0%
$-CO-NH-$

Explanation

ago group is

$RN= RN$

so correct match the group is option B is

Ethylene possess

0%
two sigma and two pi bonds
0%
two pi bonds
0%
five sigma and one pi bond
0%
four sigma and one pi bond

Explanation

Ethylene possess

$5\sigma$ and

$1\pi$ bond.
1677735_1761244_ans_bb31f5f4d49b428bb511b17823b152d7.png

Cresol has

0%
Alcoholic OH
0%
Phenolic OH
0%
$-COOH$
0%
$-CHO$

Explanation

Cresols are organic compounds in which methyl and hydroxyl groups are directly attached to the benzene ring.

Hence, option

$B$ is correct.

1626611_1763265_ans_9c0f8914af4c4ece8a4f1e9890eb4887.png

Number of

$\sigma$ and

$\pi$ bonds present in 1-buten-3-yne respectively are

0%
$7\sigma, 3\pi$
0%
$5\sigma, 2\pi$
0%
$8\sigma, 3\pi$
0%
$6\sigma, 2\pi$

Which of the following do not contain an acyl group?

0%
Acid chloride
0%
Amide
0%
Ester
0%
Ether

Explanation

Acid chloride is

$RCOCl$ , Amide is

$RCONH_2$ , Ester is

$RCOOR$ , and ether is

$ROR$ .

Acyl group has the formula

$RCO-$ . Hence, ether do not contain acyl group.

Option

$D$ is correct.

A carbon-carbon triple bond in ethyne

$(-C\equiv C-)$ consists of

0%
All $\sigma$ bonds
0%
Two $\sigma$ bonds and one $\pi-$ bonds
0%
One $\sigma$ bonds and two $\pi-$ bonds
0%
All $\pi$ bonds

Explanation

A carbon-carbon triple bonds consist of one

$\sigma$ bond and

$2\pi-$ bonds.

Which among the following is the strongest o, p-directing group

0%
$OH$
0%
$Cl$
0%
$C_{6}H_{5}$
0%
$Br$

Explanation

Meta directing deactivators< Ortho-Para directing deactivators (halogens) < Ortho -Para directing Activators

*Activator -An atom or molecule capable of increasing electron density on benzene ring either by hyper conjugation or by mesomeric effect +R .

* Halogen are dectivator due -I effect so their reactivity is less than of benzene but Ortho-Para directing due to +R mesomeric effect.

*Deactivates decreases electron density on benzene ring so electrophile can attack on meta position with difficulty.

Now the order

(Meta directing deactivators)

$-NO_2< -CN < -SO_3H < -COCH3 <-COOH < - COOCH_3 < -COH <$

( OP deactivator)

$-I < - Br < -Cl < -F <$

(OP activators )

$-H < -ph < -CH_3 < -COCH_3 < -N(CH_3)_3 < -OH < -NH_2$

Hence option A $-OH$ is a correct answer.

Alchols may be represented by the general formula:

0%
$C_{n}H_{2n+2}OH$
0%
$C_{n}H_{2n}OH$
0%
$C_{n}H_{2n+1}OH$
0%
$C_{n}H_{2n-1}OH$

Explanation

One of the H atoms of alkanes (having the molecular formula

$C_{n}H_{2n+2}$ ) is replaced by an

$-OH$ group to form alcohols.

Thus

$C_{n}H_{2n+2}-(1H)+(1-OH)\rightarrow C_{n} H_{2n+1}OH$

Option C is the correct answer.

Alkynes may be represented by the general formula:

0%
$C_{n}H_{2n+2}$
0%
$C_{n}H_{2n-2}$
0%
$C_{n}H_{2n+1}$
0%
$C_{n}H_{2n-1}$

Explanation

Alkynes are carbon compounds containing one or more

$C\equiv C$ bonds. One can generalize the formula from the examples

$C_{2}H_{2}, C_{3}H_{4}$ etc.

The decreasing order of priority for the following functional groups is:
I.

$-C\equiv N$ II.

$-CONH_{2}$ IV.

$-CHO$

0%
$(II) > (I) > (IV) > (III)$
0%
$(III) > (IV) > (I) > (II)$
0%
$(I) > (II) > (IV) > (III)$
0%
$(I) > (II) > (III) > (IV)$

Which of the following does not contain

$(-COOH)$ group?

0%
Asprin
0%
Benzoic acid
0%
Picric acid
0%
All have $(-COOH)$ group

Explanation

Picric acid is 2,4,6 trinitrophenol. It does not contain

$(COOH)$ group.
1777650_1768345_ans_4e68feac1f53478eb9f4b19b80e67230.png

Glucose has functional group

0%
Aldehydic
0%
Aldehydic and alcoholic
0%
Alcoholic
0%
Ketonic and alcoholic

Explanation

Glucose exists in a straight-chain form and in various cyclic forms. In the straight-chain form, the

$\text{functional group at C-1 is an aldehyde group.}$

$\text{The functional groups on C-2 to C-6 are alcohol groups.}$ In the cyclic form, the functional group on C-1 is a hemiacetal group
1629479_1765264_ans_957a0688adac41d6aa6156758c27ee72.jpg

1629479_1765264_ans_957a0688adac41d6aa6156758c27ee72.jpg

The decreasing order of priority for the following functional groups is:
I.

$-OH$ II.

$-C\equiv C -$ III. -OR IV.

$-NH_{2}$ .

0%
$(IV) > (I) > (II) > (III)$
0%
$(IV) > (I) > (III) > (II)$
0%
$(I) > (IV) > (III) > (II)$
0%
$(II) > (III) > (IV) > (II)$

In which of the following compounds, -OH is the functional group?

0%
Butanone
0%
Butanol
0%
Butanoic acid
0%
Butanal

Identify the correct sequence of increasing number of

$\pi$ -bonds in the structure of the following molecules :
(I)

$H_2S_2O_6$ (II)

$H_2S_2O_3$ (III)

$H_2S_2O_5$

0%
I, II and III
0%
II, I and III
0%
II, III and I
0%
I, III and II

Which of the following molecule has weakest (

$p\pi - d\pi$ ) back bonding ?

0%
$ClO_2$
0%
$N(SiH_3)_3$
0%
$SiF_4$
0%
$O(SiH_3)_2$

Explanation

$ClO_2$ has the weakest

$p\pi-d\pi$ bonding

The ratio of

$\sigma$ bond and

$\pi$ - bond in tetracyano ethylene is :

0%
$2:1$
0%
$1:1$
0%
$1:2$
0%
none of these

How many primary, secondary, tertiary and quaternary carbon atoms are present in the following compound?

0%
One primary, two secondary and one tertiary
0%
Five primary, three secondary
0%
Five primary, one secondary, one tertiary and one quaternary
0%
Four primary, two secondary and two quaternary

Explanation

$1^{0} - 5, 2^{0} - 1, 3^{0} - 1, 4^{0} - 1$

Hence option (C) is correct.

1783672_1843643_ans_b425bc14c716489fac1478bc87c92757.PNG

If the organic compound contains nitrogen and sulphur, sodium thiocyanate is always formed in the Lassaigne's test.

0%
True
0%
False

Explanation

Sodium thiocyanate is formed only when the amount of sodium is insufficient as compared to

$C, N$ and

$S$ content of the organic compound.

The decreasing order of priority for the following functional groups is:
I.

$-COOH$ II.

$-SO_{3}H$ III.

$-COOR$ IV.

$-COCl$

0%
$(IV) > (Ill) > (II) > (I)$
0%
$(I) > (II) > (lll) > (IV)$
0%
$(II) > (I) > (lll) > (IV)$
0%
$(IV) > (Ill) > (I) > (II)$

Explanation

The decreasing priority of functional groups is

$-COOH> -SO_3H> -COOR> -COCl$

$-COOH$ group is always greater priority than

$-SO_3H, -COOR, -COCl$ .

So the answer is

$B$ .

The IUPAC name of the compound is:

0%
5, 5, - diethyl - 4 - 4 - diethyl pentane
0%
3 - Ethyl - 4-4- dimethyl heptane
0%
1, 1, -Diethyl - 2, 2 - dimethyl pentane
0%
4, 4 - Dimethyl-5, 5 - diethyl pentane

The product of acid hydrolysis of P and Q can be distinguished by:

0%
Lucas Reagent
0%
2,4-DNP
0%
Fehling's Solution
0%
$\mathrm{N}\mathrm{a}\mathrm{H}\mathrm{S}\mathrm{O}_3$

Explanation

Hint: Fehling's solution can distinguish between aldehydes and ketone.

Correct Answer: Option C

Explanation:

The hydrolysis of an ester yields corresponding alcohol and acid. The alcohol is an enol, so it tautomerizes to more stable keto form.

Acid hydrolysis of

$P$ gives a ketone while acid hydrolysis of

$Q$ gives an aldehyde.

A) Lucas reagent reacts with aliphatic non-enolic alcohols only.

$2,4-DNP$ reacts with both aldehydes and ketones.

C) Fehling's solution gives Brick red precipitate with an aldehyde but not with a ketone.

$NaHSO_3$ reacts with both aldehydes and ketones.

In Lassaigne's method organic compound is fused with:

0%
sodium metal
0%
zinc dust
0%
sodium caronate and zinc dust
0%
calcium metal

Explanation

$N,S,$ halogens and phosphorus present in an organic compound are detected by Lassaigne's test, by fusing the compound with sodium

$(N_{a})$ metal, which converts the elements present in the compound from covalent to ionic form.

$Na+C+N\overset{\Delta }{\rightarrow}NaCN$

$2Na+S\overset{\Delta }{\rightarrow}Na_{2}S$

$2Na+X\overset{\Delta }{\rightarrow}NaX\ (X=Cl, \ Br \ or \ I)$

The compound (C) is:

Explanation

(2DU in A).
It may have either two

$(C=C)$ or one

$(C$

$\equiv$

$C)$ bond. Formation of three compounds on oxidation of (A) shows that it contains two

$(C=C)$ bonds.
(A)

$C_8H_{14}\Longrightarrow \underset {(B)}{CH_3CHO(two\, C \,atoms)} + \underset{(C)}{(CH_3 CPCH_2 CH_3) \text{(two C atoms)}} + \underset{(D)}{HOOC-COOH \text{(two C atoms)}}$

(2DU in A).

(B) reduces ammoniacal

$AgNO_3$ solution and gives iodoform test.
Therefore, (B) is an aldehyde and it is

$(CH_3 - CHO)$ (two

$-C-$ atom aldehyde).
(C) is a ketone containing

$\left(CH_3 - \overset{O}{\overset{||}{C}} - \right)$ group (since it gives iodoform test).
(D) is two-C dibasic acid since it gives

$CO_2$ ,

$CO,$ and

$H_2O$ with conc.

$H_2SO_4$ ,
(D) may be oxalic acid

$(HOOC - COOH)$ (two C atoms). Total C atoms in (C)must be four.

The functional group present in the following compound is:

$CH_{3} CO CH_{2} CH_{2} CH_{2} CH_{3}$

0%
ketone
0%
carboxylic acid
0%
alcohol
0%
carbonate

Explanation

The functional group present in

$CH_{3}-CO-CH_{2}-CH_{2}-CH_{2}-CH_{3}$ is the ketone group

$-CO$ .

An organic compound

$X$ give following laboratory tests.
(i) It gives white precipitate with tollen's reagent
(ii) it gives +ve iodoform test
(iii) It gives lucas test within 5 min

$X$ can be:

0%
$H-C\equiv C-(CH_2)_n - \underset{OH}{\underset{|}{C}H}-CH_2-CH_3$
0%
$CH_3-C\equiv C - (CH_2)n-\underset{OH}{\underset{|}{C}H}-CH_3$
0%
$H-C\equiv C-(CH_2)n-\underset{OH}{\underset{|}{C}H}-CH_3$
0%
none of the above

Explanation

An organic compound

$X$ give following laboratory tests.
(i) It gives a white precipitate with Tollen's reagent
(ii) It gives +ve iodoform test
(iii) It gives Lucas test within 5 min

$X$ can be the compound shown by the option

$C$ .
(i) It indicates the presence of an aldehyde group or presence of

$\displaystyle C \equiv C$ (ii) It indicates the presence of methyl keto group or a compound that can be easily oxidized to methyl keto group.
(iii) It indicates a presence of secondary alcohol.

The nodal plane in the

$\pi$ bond of ethene is located in:

0%
the molecular plane.
0%
a plane parallel to the molecular plane.
0%
a plane perpendicular to the molecular plane, which bisects the carbon-carbon sigma bond at right angle.
0%
a plane perpendicular to the molecular plane, which contains the carbon - carbon s-bond.

Explanation

The molecular plane does not contain

$\pi$ electron density. Hence, the nodal plane in

$\pi$ bond is located in the molecular plane.

$\pi$ electrons are present in a plane perpendicular to the molecular plane. The plane shown in diagram is the nodal plane in ethene.

An organic compound

$X$ having molecular mass

$60$ is found to contain

$C=20\%, H=6.67\%$ and

$N=46.67\%$ , while rest is oxygen. On heating it gives ammonia along with a solid residue. The solid residue gives violet colour with alkaline copper sulphate solution. The compound

$X$ is:

0%
$CH_3CH_2CONH_2$
0%
$CH_3NCO$
0%
$CH_3CONH_2$
0%
$(NH_2)_2CO$

Explanation

In the given compounds, only urea (

$NH_2)_2CO$ has nitrogen

$46.67$ %. Urea on heating gives a solid biuret residue which gives violet colour with alkaline

$CuSO_4$ .

Hence, option

$A$ is correct.

The IUPAC name for tertiary butyl iodide is:

0%
4-iodobutane
0%
2-iodobutane
0%
1-iodo, 3-methyl propane
0%
2-Iodo-2-methyl propane

Explanation

Hint: If an iodine group is attached to a compound then in IUPAC name $iodo$ will come.

Explanation:

The structure of tertiary butyl iodide is-

The longest chain of the compound is of $3$ carbon so the prefix is $propane$

The compound has an Iodine group and a methyl group at $C-2$ position as a substituent so it would $2-Iodo-2-methyl$ $

Its IUPAC name is – $\text{2-iodo-2-methyl propane}$ .

Correct Option: $D$

Why is functional group important?

0%
Because it is the portion of the organic molecule capable of characteristic reactions
0%
Because it is how compounds are named
0%
Because it is how you classify molecules
0%
Because it contains carbon on which all life is formed
0%
Because it is the portion of the organic molecule that does not undergo any reaction

Explanation

(A):
Functional group important because it is the portion of the organic molecule capable of characteristic reactions.
For example:

$R-OH + SOCl_2 \rightarrow R-Cl + SO_2 + HCl$
Here, the the

$-OH$ group is active for the reaction which is substituted by the

$Cl$ group in this reaction.

$CH_3-C=CH-C-OCH_3$ is named in IUPAC as:

$|$

$||$

$Cl$

$O$

0%
methyl-3-chloro-2-butenoate
0%
methy-4-chloro-2-pentanoate
0%
methoxy-3-chlorobutanol
0%
methoxy-2-chlorobutanone

Which is matched incorrectly?

0%
0%
0%
Both (a) and (b)
0%
None of the above

Which of the following group is sharp ortho and para directive?

0%
$-C_6H_5$
0%
$-OH$
0%
$-CH_3$
0%
$-Cl$

Which of following can have vacant

$p-$ orbital?

0%
$^{ \bullet }{ C{ H }_{ 3 } }$
0%
${ \overset { \oplus }{ CH } }_{ 3 }$
0%
$CH_4$
0%
Both $(B)$ and $(C)$

Explanation

Alkyl free radicals are either planar or pyramidal in structure, they have

$sp^{2}$ hybridisation with impaired

$e^{-}$ located primarily in

$p-$ orbital.

$CH_{3}$ , it has

$sp^{2}$ hybridisation with two

$e^{-}$ on each and one empty

$p-$ orbital.

$CH_{4}$ , every

$p-$ orbital is involved in hybridisation making it

$sp_{3}$ .

The compound that does not produce nitrogen has by the thermal decomposition is:

0%
$NH_4NO_2$
0%
$(NH_4)_2SO_4$
0%
$Ba(N_3)_2$
0%
$(NH_4)_2Cr_2O_7$

Explanation

$NH_4NO_2\rightarrow2H_2O+N_2$

$2(NH_4)_2SO_4\rightarrow2NH_3+H_2SO_4$

$Ba(N_3)_2\rightarrow Ba+3N_2$

$(NH_4)_2Cr_2O_7\rightarrow Cr_2O_3+N_2+4H_2O$

So, correct answer is

$(B)$ .

${ p }_{ \pi }-{ d }_{ \pi }\quad$ back bonding cannot occur in:

0%
$N{ \left( Si{ H }_{ 3 } \right) }_{ 3 }$
0%
${ PF }_{ 3 }\quad$
0%
$As{ F }_{ 3 }$
0%
${ BF }_{ 3 }$

${N}_{2}$ is estimated in organic compound by

0%
Duma's method
0%
Carius method
0%
Lassaigne's method
0%
None of these

The derived name of the above compound is:

0%
t-Butyl ethyl methyl carbinol
0%
t-Butyl ethylethanol
0%
t-Butyl ethyl methyl methanol
0%
none of these

Which of the following do not contain carbon - oxygen double bonds?

0%
Ketone
0%
Esters
0%
Acids
0%
Ethers

Explanation

Ethers do not contain carbon - oxygen double bonds. They contain

$C-O-C$ linkages.

If a mixture of methane, ammonia and oxygen is passed over pt-gauge at

$700^\circ {\text{C}}$ then the product will be:

0%
HCOOH
0%
HCN
0%
${\text{C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{H}}_{\text{2}}}$
0%
All of these

Explanation

If a mixture of methane, ammonia and oxygen is passed over pt-gauge at

$700^\circ {\text{C}}$ then the product will be HCN.

How many different functional groups are present in given compound?

0%
3
0%
4
0%
1
0%
5

Explanation

In the given compound, there are total of 4 different functional groups :

Alcohol $(-OH)$
Aldehyde $(R-CHO)$
Ketone $(R-CO-R)$
Carboxylic Acid $(-COOH)$

Thus, option B is correct.

0.5g of an organic compound was Kjeldahlised and the ammonia evolved was absorbed in 50 mL of 0.5M

$H_2SO_4$ . The residual acid required 60 mL of 0.5M NaOH.The percentage of nitrogen in the organic compound is:

0%
$14\%$
0%
$28\%$
0%
$56\%$
0%
$42\%$

Which of the following compounds gives blood red colouration when its Lassaignes extract is treated with alkali and ferric chloride?

0%
Thiourea
0%
Diphenyl sulphide
0%
Phenyl hydrazine
0%
Benzamide

The compound

$C_2FClBrI$ has:

0%
4 geometrical isomers
0%
6 geometrical isomers
0%
4 optical isomers
0%
(B) and (C) both

Explanation

From the attached fig.,

$C_{2}FClBrI$ has 6 geometrical isomers.

1185486_1071197_ans_5550c2ebe1a349948d6d30ff89e74374.png

The correct name for is :-

0%
$2-Hydroxy\ cyclopentan-1-al$
0%
$2-Formyl-1hudroxy cyclopentane$
0%
$2-Hydroxy cyclopentane carbaldehyde$
0%
$Cyclopentane -2-ol-1-al$

Explanation

Rules for Naming Hydrocarbons:-

Rule-I Selection of the longest carbon chain.

Rule-II Numbering the longest chain: Longest chain is numbered starting from a terminal carbon. Numbering is done is such a way that lowest possible number is given to the side chain or a functional group.

Rule-III Writing names of branched, saturated hydrocarbons, locants are written first in alphabetical order followed by name of parent chain.

Rule-IV Some common names are retained in $IUPAC$ system for unsubstituted hydrocarbon only.

Order of numbering:- $-CHO>-OH$ (preference)

(Refer to Image)

$\longrightarrow$ Ring is $5$ membered.

$\longrightarrow$ functional group is aldehyde.

$\longrightarrow$ at $2^{nd}$ position $-OH$ group.

So, the name of the given compound is $2-hydroxy$ $cyclopentan-1-al$ .

1008326_1070280_ans_935e1729e96f41f096bfcf42ed569f48.PNG

CBSE Questions for Class 11 Medical Chemistry Organic Chemistry Some Basic Principles And Techniques Quiz 10 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Organic Chemistry Some Basic Principles And Techniques Quiz 10 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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