43. Which of the following is true about the reaction of 3-bromo 2-methylpentane
ID: 697893 • Letter: 4
Question
43. Which of the following is true about the reaction of 3-bromo 2-methylpentane with potassium methou L. The reaction is E2 elimination with anti-periplanar substrate configuration II. The reaction is concerted WL The reaction produces the less substituted alkene due to size of the base V A favorable yield requires high concentration of 3-bromo-2-methylpentane and methanal as the solvent A 1 and ll only 8. u only O. LW, and ti only 44. Which of the following solkents may be clasuified as polar and protic? A. 1 butyl alcohol 9 Diethyl ether C. t butyl iogropyl ether E. Both A and 45. Which of the follewing organic halides wiltl undergo E1 elimination the fastest on heating with KON in A 2.2-dimethyl-1-bromopsopase 8 2.2-dimethyl-3-bromocyclsheane C. Bentyl chlende (CCHO D. 2.5-dimethl -bromabencene 45. 1 butyne has a boilne point ef 8 3'C whe a the phase of propyne at room temperature and 1 atm pressure B. Liquid C Sold D Supercrtical fl 47. The regiochemistry of hydroboration oxidation ef terminal alkynes is A Markonkov-major potuct sketone B. Non-Markovnkov-major product is an aldehyde C Folows Zaitsee's Rale D. Markovnkov-major product is an aldehyde E Non Markovnikov-major product is a ketoneExplanation / Answer
43.
Answer is option (A): I and II only.
Base is not a bulky base. So more substituted alkene will form. Dilution decreases the yield of elimination reaction.
44.
Answer is option (A) t-butyl alcohol.
Polar solvents have large dipole moments; they contain bonds between atoms with very different electronegativities, such as oxygen and hydrogen.
Protic solvents have O-H or N-H bonds. Because protic solvents can participate in hydrogen bonding, which is a powerful intermolecular force. Additionally, these O-H or N-H bonds can serve as a source of protons (H+).
45.
Answer is option (B) 2,2-dimethyl-1-bromocyclohexane
Rest four halides will not go E1 elimination reactions.
46.
Answer is option (A) Gas
Propyne is a colorless gas.
47.
Answer is option (B) Non-Markovnikov- major product is aldehyde
In hydroboration, it is observed that the regiochemistry is “anti-Markovnikov” (H ends up bound to the most substituted carbon) and the stereochemistry of the reaction is syn (both new bonds are formed on the same face of the alkene).
Again since it is a terminal alkyne it will give aldehyde.
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