Fluorine (F) has seven electrons in its outermost shell. Fluorine _____.
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(a): This is an elimination reaction. The reaction proceeds in 3 steps; Step 1: Protonation of alcohol, step 2: Loss of leaving group, and step 3: Deprotonation to form an alkene. Elimination (E1) reaction dominates than substitution (SN1) reaction because the formed HSO4- anion is a poor nucleophile and tends not to add to the carbocation intermediate formed. Hence the product formed is I, is trans.
(b) This is a dihydroxylation of the alkenes reaction. A catalytic amount of OsO4 is used along with an oxidizing agent, which oxidizes the reduced osmium(VI) into osmium(VIII) to regenerate the catalyst. The reaction proceeds in 2 steps; Step 1: Cis addition of OsO4 to double bond and form 4 member cyclic osmate ester. Step 2: Hydrolysis of the intermediate to 1.2-diol. Hence the product formed is II, is cis.
(c) This is hydroboration - oxidation reaction. The reaction proceeds through the addition of H- on the more substituted carbon and BH2 on the less substituted carbon of the double bond. The reaction is a syn addition. Hence the product formed is III.
(d) This a reduction reaction where carboxylic acid (-COOH) is reduced to 1o alcohol (-CH2OH). Hence the formed product is IV.
Stereochemistry, a subdiscipline of chemistry, involves the take a look at the relative spatial association of atoms that form the shape of molecules and their manipulation.
The observation of stereochemistry specializes in stereoisomers, which by means of definition have an equal molecular method and series of bonded atoms (constitution), however, differ within the three-dimensional orientations of their atoms in space. for that reason, it's also referred to as 3-d chemistry three-dimensionality.
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Each of the following reactions has been reported in the chemical literature. Predict the product in each case, showing stereochemistry where appropriate.
The proton which is easily abstracted in 1-Benzyl-3-propylbenzene is the proton which is present on carbon atom in between two phenyl rings, or the central carbon which is shared by two benzene rings.
This easy abstraction of proton is due to its high acidity. Remember those species are always more acidic whose conjugate base is stable. Like the acidity of carboxylic acid is due to stability of the acetate ion.
In our case the stability of conjugate base arises due to stability of negative ion due to resonance. As shown below, the negative charge can delocalize on both rings.
I have shown the resonance of negative ion on both Phenyl rings with Blue and Pink colors.<span />
This easy abstraction of proton is due to its high acidity. Remember those species are always more acidic whose conjugate base is stable. Like the acidity of carboxylic acid is due to stability of the acetate ion.
In our case the stability of conjugate base arises due to stability of negative ion due to resonance. As shown below, the negative charge can delocalize on both rings.
I have shown the resonance of negative ion on both Phenyl rings with Blue and Pink colors.<span />