Test: Reaction Mechanism Level - 3
20 Questions MCQ Test Organic Chemistry | Test: Reaction Mechanism Level - 3
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For the reaction below if the concentration of KCN is increased four times, the rate of the reaction will be:
The major product formed in the reaction of 1, 3-butadiene with bromine is at high temperature:
The products formed in the following reaction is:
The given reactant is 2-methyl-1, 3-butadiene. It is an asymmetric alkene.
- According to the reaction, 2-methyl-1, 3-butadiene is treated with one equivalent of HBr and we need to find out the major product.
- This an example of an additional reaction to asymmetric alkenes which will obey Markovnikoff’s rule.
- According to Markovnikoff’s rule, in an addition to asymmetric alkene, the major product formed will have the nucleophile or electronegative atom attached to that carbon atom which has less number of hydrogen atoms. So, protons will attack on carbon atoms having more number of hydrogen atoms.
- Now, in the given compound, there are two double bonds and only one equivalent of HBr is given. So, in the product only one double bond will be saturated and the other will remain unsaturated.
- This reaction will proceed by formation of carbocation intermediate. HBr will split to form H+�+ and Br−��−. Now, bromide ion will be the nucleophile which will attack the carbocation formed and form the product.
- The possibilities are shown as follows:
- Proton attack will happen only on C-1 and C-4 because of Markovnikoff’s rule.
- We know tertiary carbocation is more stable than secondary carbocation due to hyperconjugation effects. Therefore, even the secondary carbocation will try to show hydrogen shift and get converted to tertiary carbocation.
- Therefore, 2-Bromo-2-methylbut-3-ene will be formed as the major product in this reaction.
The major product formed in the following reaction.
The major product formed in the following reaction.
The major product obtained in the following reaction
The major product formed in the following reaction is:
NaOEt is a strong base and it will abstract the acidic proton from the alpha carbon. Dueterium is abstracted from alpha carbon because it is show more +I effect than hydrogen. After that then elimination will occur with bromide as leaving group.
Hence C
Choose the correct order of reactivity for dehydration of the given alcoho ls using concentrated sulfuric acid.
Electrophilic nitrations of the following compounds follow the trend:
The set of products formed in the following reaction is:
What would be the final major product of the following chemical reaction if it is carried out twice, one at 5 Co and the second time at 45 C0?
What is the major product if HBr (in excess) is added to H2C = CH – CH2 – OH
The reaction of sodium ethoxide with ethyl iodide to form diethyl ether is termed
What is the final product after the following reaction has gone to completion?
The major product formed in the following reaction is:
Singlet carbenes give stereospecific reactions. Option A is an example of a stereospecific reaction.
Among the bro mides I-III given below, the order of their reactivity in the SN1 reaction is:
SN1 reaction proceeds through carbocation as intermediate. Br- removed and made up carbocation in option A, B, C then we check the stability of carbonations. C is more stable because carbocation of C is aromatic, A is non aromatic and B is anti aromatic. So order of stability is:
Aromatic > Non-Aromatic > Anti Aromatic. Then more stable carbocation higher will be the reactivity and faster will be the SN1 reaction.
Hence C is correct.
The major product formed in the reaction benzoic acid with isobutylene in the presence of a catalytic amount of sulfuric acid is:
Among the following compounds, the one that undergoes deprotonation most readily in the presence of a base, to form a carbanion is:
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37 videos|91 docs|46 tests
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