Answers
The substance that is used for the industrial preparation of methyl diantilis is called the sodium borohydride (NaBH4).
Industrial preparation of methyl diantilis:
Methyl diantilis is a substance that has an elegant fragrance with a similar olfactive note to Isoeugenol.
It is used as complexing agent for vanilla, tobacco, leather and fruit accords.
It can be prepared industrially by the reduction of 3-ethoxy-4-hydroxybenzaldehyde (ethyl vanillin) to 3-ethoxy-4-hydroxybenzyl alcohol.
To reduce the aldehyde group of 3-ethoxy-4-hydroxybenzaldehyde, sodium borohydride is added in the industrial preparation process.
The sodium borohydride (NaBH4) produces a phenol functional group with other reactants leading to reduction of aldehyde group and increase in the solubility of reactants.
Therefore, the substance that is used for the industrial preparation of methyl diantilis is called the sodium borohydride (NaBH4).
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Answer:
Sodium Borohydride (NaBH₄)
Explanation:
Methyl diantilis (2-Ethoxy-4-(methoxymethyl)phenol) is a fragrance compound which smells like Vanilla. This compound is being synthesized from 3-ethoxy-4-hydroxybenzaldehyde also known as Ethyl Vanillin in two steps.
Step 1: Reduction of Aldehydic Group on Ethyl Vanillin:
The benzaldehyde derivative is treated with a mild reducing agent i.e. NaBH₄ (Sodium Borohydride). NaBH₄ is a source of Hydride (H⁻) ion and undergoes nucleophilic substitution reaction yielding 2-ethoxy-4-(hydroxymethyl)phenol.
Step 2: Etherification of 2-ethoxy-4-(hydroxymethyl)phenol:
In the second step 2-ethoxy-4-(hydroxymethyl)phenol is treated with Methanol in the presence of strong acidic polymeric resin known as Amberlyst-15-wet resulting in the formation of Methyl diantilis as shown in attached figure.
Related Questions
A student dissolves 12.g of sucrose C12H22O11 in 300.mL of a solvent with a density of 1.01/gmL . The student notices that the volume of the solvent does not change when the sucrose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits. molarity = molality =
Which of the following should be measured with a meter stick, not a tape measure?A. the length of ribbon needed to tie around a vaseB. the size of a student's waistC. the distance from the ground to the top of a rampD. the circumference of an orange
A voltaic cell with an aqueous electrolyte is based on the reaction between Cd2 (aq) and Mg(s), producing Cd(s) and Mg2 (aq). Write half-reactions for the anode and cathode and then write a balanced cell reaction. Please include the states of matter in the equations.
A mass of 34.05 g of H2O(s) at 273 K is dropped into 185 g of H2O(l) at 310. K in an insulated container at 1 bar of pressure. Calculate the temperature of the system once equi- librium has been reached. Assume that CP, m for H2O is con- stant at its values for 298 K throughout the temperature range of interest.
Answers
Answer:
4.5m/s
Explanation:
speed = distance/time
speed = 18/4
speed = 4.5m/s
Answers
Answer:
See the attached file for the structure
Explanation:
See the attached file
Answers
Answer:
(c) 18.8 g; (a) 0.798; (b) 16 mL
Explanation:
You don't give your experimental data, so I shall assume:
Mass of Al = 1.07 g
20 mL of 3 mol·L⁻¹ KOH
20 mL of 9 mol·L⁻¹ H₂SO₄
The overall equation for the reaction is
Mᵣ: 26.98 474.39
2Al + 2KOH +4H₂SO₄ + 22H₂O ⟶ 2K[Al(SO₄)₂]·12H₂O + 3H₂
m/g: 1.07
(c) Theoretical yield of alum
(i) Moles of Al
(ii) Moles of alum
(iii) Theoretical yield of alum
(a) Scaling factor for 15.0 g alum
You want a theoretical yield of 15.0 g, so you must scale down the reaction.
(b) Corrected volumes of NaOH and H₂SO₄
V = 0.798 × 20 mL = 16 mL
Answers
Answer: Cis-1-bromo-4-tert-butylcyclohexane would undergo faster elimination reaction.
Explanation:
The two primary requirements for an E-2 elimination reaction are:
1.There must be availability of β-hydrogens that is presence of hydrogen on the carbon next to the leaving group.
2.The hydrogen and leaving group must have a anti-periplanar position .
Any substrate which would follow the above two requirements can give elimination reactions.
For the structure of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane to be stable it must have the tert-butyl group in the equatorial position as it is a bulky group and at equatorial position it would not repel other groups. If it is kept on the axial position it would undergo 1,3-diaxial interaction and would destabilize the system and that structure would be unstable.
Kindly find the structures of trans-1-bromo-4-tert-butylcyclohexane and cis-1-bromo-4-tert-butylcyclohexane in attachment.
The cis- 1-bromo-4-tert-butylcyclohexane has the leaving group and β hydrogens in anti-periplanar position so they can give the E2 elimination reactions easily.
The trans-1-bromo-4-tert-butylcyclohexane does not have the leaving group and βhydrogen in anti periplanar position so they would not give elimination reaction easily.
so only the cis-1-bromo-4-tert butyl cyclohexane would give elimination reaction.
Final answer:
Trans-1-bromo-4-tert-butylcyclohexane is expected to undergo E2 elimination faster than cis-1-bromo-4-tert-butylcyclohexane due to less steric hindrance.
Explanation:
In determining the rate of E2 elimination, the trans-1-bromo-4-tert-butylcyclohexane would undergo E2 elimination faster than the cis-1-bromo-4-tert-butylcyclohexane. This is due to the larger degree of steric hindrance in the case of the cis isomer.
In trans-1-bromo-4-tert-butylcyclohexane, the bromine is at the equatorial position while the tert-butyl group is axial. It forms a structure that allows the compound to experience less steric hindrance with bromine in a more favorable position for leaving.
In comparison, cis-1-bromo-4-tert-butylcyclohexane has a bromine and tert-butyl group both at equatorial positions. This causes steric hindrance, and in turn, slows down the E2 elimination rate. Despite the more stable conformation, the bromine is not well-oriented for a leaving group in E2 elimination.
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phosphate (Ca3(PO4)2), a major component of bone.
% Ca
%P
% 0
Answers
Answer:
38.7%
41.3%
20%
Explanation:
The percentage composition helps to know the what percent of the total mass of a compound is made up of each of the constituent elements or groups.
To solve this problem:
- find the formula mass by adding the atomic masses of the atoms that makes up the compound.
- place the mass contribution of the element or group to the formula mas and multiply by 100;
Compound:
Ca₃(PO₄)₂
Formula mass = 3(40) + 2[31 + 4(16)]
= 120 + 2(95)
= 120 + 190
= 310
%C = x 100 = 38.7%
%P = x 100 = 41.3%
%O = x 200 = 20%
Answers
Answer:
cilia and flagella
Explanation:
In prokaryotic species , cilia are present , and in eukaryotic species , flagella is present .
Cilia and flagella both have same function , i.e. , to enable the movement of the cell , along with the movement of some substance and direct the flow of these substance along the tracts.
Cilia and flagella are composed of basal bodies.
Hence , from the given statement of the question,
The correct term is cilia and flagella .
Final answer:
Cilia are the structures that help move substances across the surface of a tract. They are primarily found in certain types of cells such as those in the respiratory tract and the oviducts.
Explanation:
The structures that assist in moving substances across a tract surface are primarily called cilia. In biology, cilia are tiny hair-like structures that line certain types of cells, especially those in the respiratory tract and the oviducts. They work much like oars on a boat, moving in coordinated waves to propel substances (like mucus or egg cells) along the surface of the tract they occupy. They are responsible in movement of substances across tract.
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