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CBSE Questions for Class 12 Engineering Chemistry Aldehydes,Ketones And Carboxylic Acids Quiz 12 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Aldehydes,Ketones And Carboxylic Acids
Quiz 12
The reaction of $$C_6H_5CH=CHCHO$$ with $$LiAlH_4$$ gives.
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$$C_6H_5CH_2CH_2CH_2OH$$
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$$C_6H_5CH=CHCH_2OH$$
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$$C_6H_5CH_2CH_2CHO$$
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$$C_6H_5CH_2CHOHCH_3$$
Explanation
$$LiAlH_4$$ does not reduce alkene, but when the alkene is in conjugation with aldehyde and phenyl group is attached to it, then alkene also get reduced.
In amine formation, $$p^H$$ must maintain
$$R - C - H \longrightarrow R - CH = N - R $$
At high $$p^H$$, there wont be enough acid to protanate the $$OH$$ in the intermediate to allow for removal as $$H_2O$$. At low $$p^H$$, most of the amine reactant will be tied up as its ammonium conjugate acid and will become non-nucleophilic.
Compound(x) in the given reaction is:
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$$Ph-\overset{O}{\overset{||}{C}}-CH_3$$
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$$Ph-\overset{O}{\overset{||}{C}}-H$$
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$$Ph-CH_2-\overset{O}{\overset{||}{C}}-H$$
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$$Ph-CH_2-\overset{O}{\overset{||}{C}}-CH_3$$
Explanation
(Refer to Image)
So, the reactant $$X$$ is $$Ph-CHO$$ .
Complete the reactions with appropriate products.
(i) $$CH_3CHO+NH_2OH \rightarrow X$$
(ii) $$CH_2=CH_2+PdCl_2+H_2O \xrightarrow{CuCl_2}Y $$
(iii) $$CH_3CH_2CH_2CH_2OH \xrightarrow{CrO_3-H_2SO_4}Z $$
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$$X = CH_3CH=NOH, \,\, Y = CH_3CHO, \,\, Z = CH_3CH_2CH_2COOH$$
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$$X = CH_3CH_2NH_2, \,\, Y= CH_3CH_2CHO, \,\, Z = CH_3CH_2CH_2COOH$$
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$$X = CH_3CONH_2, \,\, Y= HCHO, \,\, Z= CH_3COCH_3$$
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$$X = CH_3C\equiv N, \,\, Y = CH_3CHO,\,\, Z = HCOOH$$
Explanation
When butanol undergo reaction with $$CrO_3 - H_2SO_4$$ in presence of sulphuric acid, it poroduce batanoic acid.
Hence, the products are
$$X \rightarrow CH_3 = NOH$$
$$Y \rightarrow CH_3 - CHO$$
$$Z \rightarrow$$ $$CH_3 CH_2 CH_2 COOH$$
Consider the following sequence of reactions. The final product (B) is?
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Explanation
In the above reacation $$HIO_4$$ cleanes 1,2 diole into two carbonyl compounds . i,e the reaction is selective for 1 ,2 diole. this reaction occurs via the formation of a cyclic periodic ester. It can be used as a functional group test for 1,2 diole also the products are determined by the substituent on the diole.
Product (x) in this reaction is:
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Possible product of the following reaction is:
$$CH_3CHO + CH_3 CH_2 CHO \xrightarrow[(ii)\Delta]{(i)NaOH}$$
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$$CH_3CH_2CH_2OH$$
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$$CH_3CH = \overset{CH_3}{\overset{|}{C}} - CHO$$
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$$CH_3CH_2CH = CH - CH_2OH$$
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$$CH_2 = CH - CHO$$
The IUPAC name of the compound is:
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$$2-ethyl-3-methylcyclohexa-1, 3-diene$$
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$$2, 5-dimethylhepta-2, 6-dienoic$$ acid
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$$3-bromo-2-methylbutanoic$$ acid
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$$3-bromo-2, 3 dimethylbutanoic$$ acid
The correct IUPAC name of the compound is:
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$$3$$-amino-$$6$$-bromocyclohexane-$$1$$-carboxylicacid
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$$2$$-bromo-$$5$$-aminocyclohexane-$$1$$-carboxylic acid
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$$5$$-amino-$$2$$-bromocyclohexane-$$1$$-carboxylic acid
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$$2$$-bromo-$$5$$-carboxycyclohexanamine
Which of the following will not form acetyl chloride with $$ PCl_{5} $$?
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MeCOOH
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MeCOOMe
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MeCOOCOMe
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$$ MeCONH_{2} $$
Cyanohydrin of which of the following gives lactic acid on hydrolysis ?
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$${ CH }_{ 3 }{ COCH }_{ 3 }$$
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$${ CH }_{ 3 }{ CHO }$$
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$${ C }_{ 6 }{ H }_{ 5 }{ CH }_{ 2 }CHO$$
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$$HCHO$$
Explanation
Cyano hydrin of acetone upon hydrolysis gives the lactic acid compound.it is a type nucleophilic addition reaction followed by elimination of water molecule.
Product (C) of the reaction is:
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Among the following which one is least soluble in water?
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$$HCHO$$
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$$CH_{3}CHO$$
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$$CH_{3}CH_{2}CH_{2}CHO$$
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$$CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}CHO$$
The sodium salt of an organic acid 'X' produces effervescence with conc. $$H_2SO_{4}$$. 'X' reacts with the acidified aqueous $$CaCl_2$$ solution to give a white precipitate which decolorizes acidic solution of $$KMnO_4$$. 'X' is:
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$$C_6H_5COONa$$
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$$HCOONa$$
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$$CH_3COONa$$
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$$Na_2C_2O_4$$
Explanation
undefined
$$\underrightarrow { NaOH\left( aq \right) }$$
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Identify the products(s) obtained in the following reaction
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$$CH_3COOH$$
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$$CH_3CH_2CH_2COOH$$
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$$CH_3CH_2COOH$$
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All of these
When a nucleophile attacks a carbonyl group to form an intermediate, the hybridisation of the carbon atom changes from
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$$sp^{3}$$ to $$sp^{2}$$
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$$sp^{2}$$ to $$sp$$
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$$sp$$ to $$sp^{2}$$
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$$sp^{2}$$ to $$sp^{3}$$
Which of the following statements is incorrect regarding the reaction?
$$CH_3CHO+[Ag(NH_3)_2]^{\oplus} + \bar{OH} \rightarrow CH+3COO^-+Ag$$
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The equivalent weight of $$CH_3CHO$$ is 22
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Three moles of $$OH^-$$ are required in the above reaction
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$$CH_3CHO$$ is an oxidising agent
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Reduction of $$[Ag(NH_3)_2]^{\oplus}occurs$$
Explanation
Oxidation:
$$CH_3CHO+3\bar{OH} \rightarrow CH_3COO^-+2e+2H_2O$$ ...(i)
Reduction :
$$\{Ag(NH_3)_2]^++e^{\oplus}\rightarrow Ag+2NH_3\}\times 2$$ ...(ii)
$$\therefore CH_3CHO+[Ag(NH_3)_2]^{\oplus}+3\bar{OH}\rightarrow CH_3COO^-+2H_2O+2Ag+4NH_3$$
Molecular mass of $$CH_3CHO=44$$
Equivalent weight $$=\dfrac{MW}{n-factor}=\dfrac{44}{2}=22$$
Therefore, acetaldehyde reducing agent
Hence, the correct option is $$\text{C}$$
$$C_nH_{2n-1}COOH$$ is the general formula for:
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saturated fatty acids
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unsaturated fatty acids
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simple lipids
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phospholipids
Explanation
$$C_nH_{2n-1}COOH$$ is a general formula for unsaturated fatty acids.
For example: $$CH_2=CH-COOH$$
Propenoic acid $$C_2H_3COOH$$
$$C_nH_{2n+1}COOH$$ is general formula for saturated fatty acids.
For example: $$CH_3COOH$$ is Ethanoic acid.
Reduction of ketones cannot be carried out with which of the following reagents?
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Zinc amalgam and concentrated HCl
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Hydrazine and KOH in ethylene glycol
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Sodium borohydride or Lithium Aluminium hydride
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Hydrogen in presence of palladium in Barrium sulphate and quinoline
Explanation
Reduction of ketones cannot be carried out with hydrogen in presence of palladium in Barrium sulphate and quinoline. This reagent is Lindlar's catalyst and is used for selective reduction of alkyne to trans alkene.
In Clemmensen's reduction, zinc amalgam and concentrated HCl converts carbonyl group to methylene group.
in Wolf Kishner's reduction, Hydrazine and KOH in ethylene glycol converts carbonyl group to methylene group.
Sodium borohydride or Lithium Aluminium hydride reduces keto groups to alcohols.
One girl student uses the liquid compound to remove nail paint from her nail. The liquid compound removes the nail paint from her nail. Give the name of this compound.
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Acetic acid
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Formalin
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Ethanol
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Propanone
Explanation
Propanone also called acetone is used to remove nail paint from the nail.
Acetone is a volatile, flammable and colorless liquid that is miscible with water. On the other hand, nail polish remover is an organic solvent that may include coloring, scents, oils, and solvents.
Option D is correct.
IUPAC name of valeric acid is:
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Propanoic acid
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Butanoic acid
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Ethanoic acid
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Pentanoic acid
Explanation
The common name of Pentanoic acid is valeric acid. Valeric acid, or pentanoic acid, is a straight-chain alkyl carboxylic acid with the chemical formula $$C_5H_{10}O_2$$.
Which of the following is crossed aldol product in the reaction?
$$CH_3CHO+CH_3CH_2CHO\xrightarrow[25^oC]{OH^-}$$ $$'P'$$
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Explanation
Ethanal is more electrophilic than propanal. So, enolate of propanal will attack as nucleophile on ethanal.
The product formed is $$\beta- hydroxy\ aldehyde$$ , given in option C.
State the oxidation number of carbonyl carbon in methanal and methanoic acid respectively:
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$$0$$ and $$0$$
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$$0$$ and $$+2$$
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$$+1$$ and $$+2$$
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$$+1$$ and $$+3$$
Explanation
Let the oxidation number of carbonyl carbon in methanal $$(HCHO)$$ and methanoic acid $$(HCOOH)$$ is $$x$$ and $$y$$ .respectively
In $$HCHO$$,
$$2(+1)+x+(-2)=0$$
$$2+x-2=0$$
$$x=0$$
In $$HCOOH$$
$$2(+1)+y+2(-2)=0$$
$$y=2$$
The general formula $${ C }_{ n }{ H }_{ 2n }{ O }_{ 2 }$$ could be for open chain
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Diketones
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Carboxylic acid
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Diols
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Diadehydes
Explanation
$${ C }_{ n }{ H }_{ 2n }{ O }_{ 2 }$$ is the general formula for open chain carboxylic acid and ester e.g if $$n=3\Rightarrow { C }_{ 3 }{ H }_{ 6 }{ O }_{ 2 }$$
Ester-$${ CH }_{ 3 }COO{ CH }_{ 3 }$$
Acid-$${ CH }_{ 3 }{ CH }_{ 2 }COOH$$
IUPAC name of the above compound is:
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bicyclo[3,2,1]-octan-2-oic acid
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bicyclo-[3,2,1]-octane-8-carboxylic acid
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bicyclo-[3,2,1]-octan-8-oic acid
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none of these
Explanation
The IUPAC Name is
bicyclo-[3,2,1]-octane-8-carboxylic acid
Which of the following statements is not true for acetaldehyde ?
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It will undergo an aldol condensation reaction.
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It will undergo a Cannizzaro reaction.
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It will undergo a haloform reaction
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Enolisation is catalysed by acid or base.
Explanation
Acetaldehyde contains $$\alpha-H$$, so it undergoes aldol condensation rather than Cannizzaro reaction.
It undergoes the haloform reaction as it contains $$-COCH_3$$ group.
As it contains the alpha hydrogen it undergoes enolisation.
What is the oxidation number of carbonyl carbon in acetophenone?
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$$+3$$
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$$+1$$
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$$+2$$
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Zero
Explanation
Thus, oxidation number of carbonyl carbon in acetophenone is $$+2$$. It is attach to two C atom of same EN and 1 O atom which has higher EN value via double bond.
When a nucelophile encounters a ketone the site of attack is:
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The carbon atom of the carbonyl
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The oxygen atom of the carbonyl
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Both the carbon and oxygen atoms with equal probability
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No attack occurs-ketones do not react with nucleophiles
Explanation
Nucleophile attacks on the carbon atom of the carbonyl group as shown in the figure.
$$A\xrightarrow [ ]{ HCN } B\xrightarrow [ ]{ { H }_{ 3 }{ O }^{ + } } Lactic \ acid$$
Identify $$A$$.
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$$HCHO$$
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$${ CH }_{ 3 }CHO$$
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$${ C }_{ 6 }{ H }_{ 5 }CHO$$
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$${ CH }_{ 3 }CO{ CH }_{ 3 }$$
Explanation
answer is (b)
Artificial sweetening agent is prepared from the strategy. Identify the structure of final product.
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Explanation
Artificial sweetening agent is prepared from the strategy. The structure of final product is represented by option (A). The final product is saccharine. It goes directly through the human digestive system without being digested. It is 550 times more sweet than cane sugar.
Among the following acids which has the lowest $$ pk_a $$ value?
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$$CH_3 CH_2 COOH$$
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$$ (CH_3)_2 CHCOOH$$
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$$ HCOOH $$
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$$ CH_3 COOH $$
Explanation
Lowest $$pk_a\rightarrow$$ Highest $$k_a\rightarrow$$ High acididty.
As $$R$$ increases, acidity decreases in carboxylic acid due to increase in electron density on $$O-H$$ bond.
Where, $$R$$ is length of alkyl chain.
So, $$HCOOH$$ is more acidic and least value of $$pk_a$$
Hence, option $$C$$ is correct.
What is the final product C?
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Which oxidised product is obtained when benzene diazonium chloride reacts with ethanol?
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Acetaldehyde
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Phenol
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Benzaldehyde
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Benzene
Explanation
The oxidized product of benzene diazonium chloride reacts with ethanol gives acetaldehyde.
Which of the following acid does not have $$-COOH$$ gourp?
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Ehtanoic acid
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Picric acid
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Benzoic acid
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Salicylic acid
Explanation
Picric acid doesn't contain carboxylic acid. It is stronger than acetic acid also.
Consider the following sequence of reactions.
The products $$(A)$$ and
(
B
)
(B)
are:
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Explanation
Ketone gives hydroxylamine products on reaction with Hydroxylamine and heating/oxime.
The product of given reaction is:
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Find the major product.
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Which of the following is the correct reactant for the given reaction?
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None of these
In the given transformation, which of the following is the most appropriate reagent?
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$${ NH }_{ 2 }{ NH }_{ 2 },\ OH$$
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$$Zn\ -Hg/HCl$$
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$$Na,\ liq.\ { NH }_{ 3 }$$
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$$NaB{ H }_{ 4 }$$
Explanation
In the given transformation, $$NH_2 - NH_2, OH$$ is the most appropriate regent. Because in this regent the medium is basic so it will not have any effect on the $$-OH$$ group in the regent.
But in $$Zn - Hg/ HCl$$, the medium is acidic
$$\therefore$$ after the conversion of $$-\overset{O}{\overset{||}{C}} - $$ into $$-CH_2$$ group.
$$NaBH_4$$ converts $$-\overset{O}{\overset{||}{C}}-$$ group into $$-\overset{OH}{\overset{|}{C} -} $$. $$Na, liq$$.
$$NH_3$$ converts alkyne into alkene.
$$\therefore$$ Ans is (A) $$NH_2 - NH_2, OH$$
What will be the most reactive among the following carbonyl compounds with Grignard's reagent?
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Explanation
Addition of the bulky group to the carbonyl carbon atom decreases the reactivity of the carbonyl carbon atom. Due to the positive Inductive effect of two bulky substituents reduces the electrophilicity of carbonyl carbon.
The bulkier group reduces the approach of the nucleophile to the carbonyl carbon.
Hence the compound is given in option A i.e., Formaldehyde will be most reactive with Grignard's reagent.
In which of the following compounds the prefix for side chain attached to the parent chain\ring is correct?
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Explanation
Reffer image.
Prefix $$\rightarrow $$ methoxycarbonyl
The major product of the following reaction sequence is ?
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$$B$$ is ?
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Explanation
Refer to image
Alkene on reacting with acidic $$KMnO_4$$, diols are not formed. Alkenes are claved. The intially formed dids are themselves quite easily oxidised by $$Mn^{+7}$$ ions gives carboxylic acid.
Which of the following carbonyl compounds is most polar?
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$$C_2H_5-{\overset{O}{\overset{||}{C}}}-C_2H_5 $$
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$$CH_3-{\overset{O}{\overset{||}{C}}}-CH_3 $$
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$$CH_3-{\overset{O}{\overset{||}{C}}}-H $$
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$$H-{\overset{O}{\overset{||}{C}}}-H $$
Explanation
More the electrophilicity of carboxyl-carbon more will be the positive charge on carbon.
$$\therefore$$ More will be the negative charge on oxygen
$$\therefore$$ More will be polarity.
Tartaric acid is a colourless, crystalline acid found in fruit juices, like that of grapes. Some of its uses are mentioned here. Pick the false one:
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In photography
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In cosmetics
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In dyeing of cloth
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In medicines
Explanation
$$Tartaric\quad acid$$
Tartaric acid is used in pickles, juices, medicine. Tartaric acid occur naturally in plants like grapes, apricot, apple, bananas, tamarinds.
$$\alpha$$-Hydroxypropanoic acid can be prepared from ethanal by following the steps given in the sequence.
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Treat with $$HCN$$ followed by acidic hydrolysis
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Treat with $$NaHSO_4$$ followed by reaction with $$Na_2CO_3$$
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Treat with $$H_2SO_4$$ followed by hydrolysis
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Treat with $$K_2Cr_2O_7$$ in presence of sulphuric acid
Explanation
From the above reaction treatment of Ethanal with $$HCN$$ followed by acidic hydrolysis gives $$\alpha-$$Hydroxy propanoic acid.
Which among the following is most reactive to give nucleophilic addition?
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$$FCH_2CHO$$
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$$ClCH_2CHO$$
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$$BrCH_2CHO$$
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$$ICH_2CHO$$
Explanation
$$FCH_2CHO$$ is most reactive towards nucleophilic addition since presence of most electronegative $$F$$ withdraws electron from carbon of carbonyl group making it more polar.
Choose the correct statement regarding the physical properties of carbonyl compound.
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All aldehydes are insoluble in benzene
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Higher aldehydes are more fragrant
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n-Butane has more boiling point than acetone
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Methanal and propanone are immiscible with water in all proportions
Explanation
Higher Aldehydes are soluble ib Benzene.
lower Aldehydea are pungent but higher Aldehydes are fragrant.
acetone is more polar than n-Butane and weight is comparable ,so acetone has higher boiling point.
Methanal is soluble in water.
Which of the following reactions does not occur?
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Explanation
For oxidation of hydrocarbon, at least one
α
α
hydrogen is necessary otherwise no oxidation will occur.
Ionic species are stabilised by the dispersal of charge, which of the following carboxylate ions is the most positive charge?
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$$CH_{ 3 }-\overset { \underset { || }{ \displaystyle \text {O} } }{ C } -\overset { - }{ O } $$
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$$Cl-CH_{ 2 }-\overset { \underset { || }{ \displaystyle \text {O} } }{ C } -\overset { - }{ O } $$
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$$F-CH_{ 2 }-\overset { \underset { || }{ \displaystyle \text {O} } }{ C } -\overset { - }{ O } $$
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$$F-\underset{\underset{\displaystyle \text {F}}{|}}{C}H_{ 2 }-\overset { \underset { || }{ \displaystyle \text {O} } }{ C } -\overset { - }{ O } $$
Explanation
As $$F$$ is the most electronegative and in (D) structure, there are two $$F$$ atoms, therefore, dispersal of negative charge is maximum, hence it is the most stable.
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