Alkenes on reaction with meta-chloroperoxybenzoic acid (MCPBA ) produces epoxides. When styrene is reacted with mCPBA it gives 2-phenyloxirane as shown below,
Answer:
23.71J is the work that the gas do.
Explanation:
The work that a gas do under isobaric conditions follows the formula:
W = P*ΔV
<em>Where W is work in atmL, P is the pressure and ΔV is final volume -Initial volume In Liters</em>
Replacing with the values of the problem:
W = P*ΔV
W = 0.600atm*(0.44000L - 0.0500L)
W = 0.234atmL
In Joules (1atmL = 101.325J):
0.234atmL × (101.325J / 1 atmL) =
<h3>23.71J is the work that the gas do.</h3>
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Answer:
1) positive
2) carbocation
3) most stable
4) faster
Explanation:
A common test for the presence of alcohols can be achieved using the Lucas reagent. Lucas reagent is a mixture of concentrated hydrochloric acid and zinc chloride.
The reaction of Lucas reagent reacts with alcohols leading to the formation of an alkyl chloride. Since the reaction proceeds via a carbocation mechanism, tertiary alcohols give an immediate reaction. Once a tertiary alcohol is mixed with Lucas reagent, the solution turns cloudy almost immediately indicating an instant positive reaction.
Secondary alcohols may turn cloudy within five minutes of mixing the solutions. Primary alcohols do not significantly react with Lucas reagent obviously because they do not form stable carbocations.
Therefore we can use the Lucas reagent to distinguish between primary, secondary and tertiary alcohols.
Answer:
a) ammonium ion
b) amide ion
Explanation:
The order of decreasing bond angles of the three nitrogen species; ammonium ion, ammonia and amide ion is NH4+ >NH3> NH2-. Next we need to rationalize this order of decreasing bond angles from the valence shell electron pair repulsion (VSEPR) theory perspective.
First we must realize that all three nitrogen species contain a central sp3 hybridized carbon atom. This means that a tetrahedral geometry is ideally expected. Recall that the presence of lone pairs distorts molecular structures from the expected geometry based on VSEPR theory.
The amide ion contains two lone pairs of electrons. Remember that the presence of lone pairs causes greater repulsion than bond pairs on the outermost shell of the central atom. Hence, the amide ion has the least H-N-H bond angle of about 105°.
The ammonia molecule contains one lone pair, the repulsion caused by one lone pair is definitely bless than that caused by two lone pairs of electrons hence the bond angle of the H-N-H bond in ammonia is 107°.
The ammonium ion contains four bond pairs and no lone pair of electrons on the outermost nitrogen atom. Hence we expect a perfect tetrahedron with bond angle of 109°.
Answer:
it is actually b because i did this i picked b and got it right
Explanation:
