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Maru [420]
3 years ago
14

Construct a three-step synthesis of trans-2-pentene from acetylene by dragging the appropriate formulas into the bins. Note that

each bin will hold only one item, and not all of the given reagents or structures will be used.

Biology
1 answer:
lozanna [386]3 years ago
5 0

Answer:

A. Production of higher order terminal alkyne on reaction with sodium acetylides with haloalkane.

B. It includes the further alkylation of terminal alkynes to higher order nonterminal alkynes and

C. The production of trans-pent-2-ene by the use of the dissolving reduction method.

Explanation:

Concepts and reason

The basic idea of this case study is on the conversion of an alkyne to an alkene. Alkynes are unsaturated and can go through addition reactions. The essential chemical processes of alkynes involves the electrophilic addition reactions, nucleophilic addition reactions , reactions as a result of acetylenic hydrogen, polymerisation, isomerization and oxidation reactions.

Fundamentals

The conversion of an alkyne to alkene involves two basic processes which as to do with the production of higher alkynes from lower alkynes with the next phase involving the dissolved metal reduction of the higher order alkyne so as to form an alkene.

Conversion of acetylene (ethyne) to one higher order alkyne as shown in the first attached image.

Terminal alkynes combine with sodamide to produce acetylides that reacts with the haloalkanes (In this case, we using methyl bromide) to form higher order alkynes.

The reaction is as well represented in the second attached image.

Alkylation of terminal alkyne leads to the production of higher nonterminal alkyne (which is propyne) that is converted to higher order nonterminal Pent-2-yne.

The reaction is shown in the third attached image.

The entire or total hydrogenation of alkynes produces a corresponding alkane by passing hydrogen in the presence of metal catalysts. Partial hydrogenation of alkynes could lead to the development of cis- or trans- alkenes which is reliant on the reagent employed.

The right reagents and products in the total reaction is duly represented in the fourth attached image.

The Synthesis of trans-pent-2-yne from ethyne involves basically a three-step synthesis which includes

A. Production of higher order terminal alkyne on reaction with sodium acetylides with haloalkane.

B. It includes the further alkylation of terminal alkynes to higher order nonterminal alkynes and

C. The production of trans-pent-2-ene by the use of the dissolving reduction method.

• The last attached image has the answers being put in the adequate bin.

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The following F2 results occur from a dihybrid cross (AaBb x AaBb): purple: A_B_ 9/16 white: aaB_ 3/16 white: A_bb 3/16 white: a
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Answer:

b. 3 (whilte) : 1(purple)

Explanation:

The first dihybrid crossing is between AaBb x AaBb

Now, if the double heterozygous traits self crossed, we have the following gametes shown below for the F₂ crossing.

AaBb = (AB, Ab, aB, ab)

                    AB                     Ab                     aB                     ab

AB                AABB                AABb                AaBB                AaBb

Ab                AABb                AAbb                AaBb                Aabb

aB                AaBB                 AaBb                aaBB                aaBb

ab                AaBb                 Aabb                 aaBb                aabb

We were being told that the results are;

purple: A_B_ 9/16

white: aaB_ 3/16

white: A_bb 3/16

white: aabb 1/16

From above, we can see that the same is true;

For Purple color; we Have (AABB and AaBb) since A is dominant to a and B is dominant to b.

∴ From the above punnet square; we have 9 purple colors which are:

(AABB, AABb, AaBB, AaBb, AABb, AaBb, AaBB, AaBb, AaBb) = 9/16

white: aaB_ ( since B is dominant to b, it can be either BB or Bb)

= (aaBB, aaBb, aaBb) = 3/16

white: A_bb ( since A is dominant to a, it can be either AA or Aa)

= (AAbb, Aabb, Aabb) = 3/16

white: aabb i.e homozygous recessive = (aabb) = 1/16

Furthermore, the question goes further by saying:

If a double heterozygote (AaBb) is crossed with a fully recessive organism (aabb), what phenotypic ratio is expected in the offspring?

If AaBb self crossed, we have:  (AB, Ab, aB, ab)

If aabb self crossed, we have: (ab, ab, ab, ab)

                    AB                     Ab                     aB                     ab

ab                AaBb                Aabb                aaBb                 aabb

ab                AaBb                Aabb                aaBb                 aabb

ab                AaBb                 Aabb               aaBb                 aabb

ab                AaBb                 Aabb                aaBb                aabb

AaBb ( purple)

= 4/16

= 1/4

= 0.25

= 25%

Aabb (white)

= 4/16

= 1/4

= 0.25

= 25%

aaBb (white)

= 4/16

= 1/4

= 0.25

= 25%

aabb (white)

= 4/16

= 1/4

= 0.25

= 25%

Now the proportion of white to purple =  (25%+25%+25%): 25%

=  75%:25%

= 3(white):1(purple)

We can therefore conclude that the expected phenotypic ratio in the cross between a double heterozygote (AaBb) with a fully recessive organism (aabb) yeids;

3 (whilte) : 1(purple)

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