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Sudip C.
Certified Chemistry Teacher
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Physical Chemistry
TutorMe
Question:

$\textup{Reactions that are catalysed by metals are often zero order after a certain amount of reactant(s) are present. Explain }$

Sudip C.

$\textup{The zero order reactions are apparently independent of the reactant concentration. So it does not vary with increasing or decreasing reactants concentrations. In a metal catalyzed reaction, if some amount of reactants is present then the metal surface is completely covered with the reactant molecules and under this condition the rate is a constant because it is controlled by what happens on the metal surface rather than the excess concentration. As a result the reaction reverts back to zero order reaction since the reactants are no longer used for the reaction. If they are used in the reaction then the reaction becomes first or second order. Since excess reactants are remained unused so it is zero order. }$

Organic Chemistry
TutorMe
Question:

$\textup{Dissolving metal reductions reduce alkynes to trans alkenes. Why do these reactions fail with a terminal alkyne? }$

Sudip C.

$\textup{Reduction of alkynes to trans-1,2-disubstituted alkenes is possible using dissolved metal reduction methods like sodium metal in liquid ammonia. But attempts to partially reduced terminal alkynes to alkene by this method fail. This is because alkyne reacts in the first step to produce alkynyl sodium species (sodium acetylide). Unlike substituted alkynes, terminal alkyne does not form any trans or cis radical. Sodium acetylide is formed exclusively. Since negative charge resides on the alkynyl carbon atom which is sp-hybridized, hence sodium acetylide resists reduction. Hence the reduction of terminal alkyne to alkene fails by using dissolved metal reduction method }$

Chemistry
TutorMe
Question:

$\textup{Explain why the Lewis acid character of boron trihalides is found to follow the order BI$_{3}>$BBr$_{3}>$BCl$_{3}>$BF$_{3}$}$

Sudip C.

$\textup{On the basis of the relative electronegativities of the halogens, one might expect the B atom in BF$_{3}$to be most electrons deficient and hence most acidic. However, the compound involve extensive$\pi{}$-interaction from a filled p orbital on the halogen and an empty p-orbital on B. The smaller F-atoms form the most stable p-p$\pi{}$-bond by efficient match of energy and size of the orbitals. Formation of an adduct with a base converts the geometry of bonds around boron from planar to pyramidal, thus rupturing the$\pi{}$-bonds. The Lewis acidity therefore increases with decrease in the extent of$\pi{}$-conjugation, as the halogen atom increases in size giving poorer overlap, i.e., from BF$_{3}$to BI$_{3}$. Therefore the order is BF$_{3}<$BCl$_{3}<$BBr$_{3}<$BI$_{3}$}$

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