
Concept explainers
Interpretation:
The retro synthesis analysis for the given compound is to be developed, the reactions are to be written for the synthesis of this compound, the synthetic intermediates show IR absorption that would result from retrosynthetic analyses and its 3-d structure of major product is to be drawn.
Concept Introduction:
▸ Electrophiles are electron deficient species which has positive or partially positive charge. Lewis acids are electrophiles which accept electron pair.
▸ Nucleophiles are electron rich species which has negative or partially negative charge. Lewis bases are nucleophiles which donate electron pair.
▸ Substitution reaction: A reaction in which one of the hydrogens of a hydrocarbon or a functional group is substituted by any other functional group.
▸ Elimination reaction: A reaction in which two substituent groups are detached and a double bond is formed.
▸ Addition reaction: A reaction in which unsaturated bonds are converted to saturated molecules by addition of molecules.
▸ Retrosynthesis: A process by which the reaction used to form target product is deduced by determining the immediate reactant used to produce it and then repeating the step again to determine other precursors.
▸ The molecules which are non-superimposable or not identical with their mirror images are known as chiral molecules.
▸ A pair of two mirror images which are non-identical is known as enantiomers which are optically active.
▸ The objects or molecules which are superimposable with their mirror images are achiral objects or molecules and these objects have a centre of symmetry or plane of symmetry.
▸ The achiral compounds in which plane of symmetry is present internally and consists of chiral centres are known as meso compounds but they are optically inactive.
▸ The stereoisomers which are non-superimposable on each other and not mirror images of each other are known as diastereomers.
▸ Chiral molecules are capable of rotating plane polarized light
▸ The molecules which are superimposable or identical with their mirror images are known as achiral molecules, and achiral molecules are not capable of rotating the plane-polarised light.
▸ The reaction in which hydrogen is added to the compound in the presence of catalyst is known as hydrogenation.
▸ The number of moles of hydrogen absorbed will be equal to the number of double bonds.
▸ Infrared spectroscopy is a simple, instrumental technique, which helps to determine the presence of various
▸ It depends on the interactions of atoms or molecules with the
▸ Sodium amide is a strong base and it helps in the formation of acetylide that can be converted into bigger
▸ Reduction is a process in which hydrogen atoms are added to a compound. Usual reagent used in reduction process is
▸ Chair conformations: It is the most stable conformation, which accurately shows the spatial arrangement of atoms.
▸ Equatorial bonds are parallel to the average plane of the ring, while axial bonds are perpendicular to the average plane of the ring.
▸ The conformation having bonds at the equatorial positions are more stable than those with bonds at the axial position.
▸ On flipping the cyclohexane ring, axial bonds become equatorial bonds and equatorial bonds becomes axial bond.
▸ Bulkier group acquires equatorial positions to form stable conformer due to steric factors.
▸ In double bond or cyclic compounds, if two same functional groups are present on the same side of the double bond or cyclic compound, the given compound can be labeled as cis.
▸ If the two functional groups are present on the different sides of the double bond or cyclic compound, the given compound can be labeled as Trans.
▸ Cis-trans isomerism exists in the compounds in which similar groups are present on the adjacent carbon atoms.

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Chapter 7 Solutions
ORGANIC CHEMISTRY-WILEYPLUS ACCESS PKG.
- A J то گای ه +0 Also calculate the amount of starting materials chlorobenzaldehyde and p-chloroacetophenone required to prepare 400 mg of the given chalcone product 1, 3-bis(4-chlorophenyl)prop-2-en-1-one molar mass ok 1,3-bis(4-Chlorophenyl) prop-2-en-1-one = 277.1591m01 number of moles= 0.400/277.15 = 0.00144 moles 2 x 0.00 144=0.00288 moves arams of acetophenone = 0.00144 X 120.16 = 0.1739 0.1739x2=0.3469 grams of benzaldehyde = 0.00144X106.12=0.1539 0.1539x2 = 0.3069 Starting materials: 0.3469 Ox acetophenone, 0.3069 of benzaldehyde 3arrow_forward1. Answer the questions about the following reaction: (a) Draw in the arrows that can be used make this reaction occur and draw in the product of substitution in this reaction. Be sure to include any relevant stereochemistry in the product structure. + SK F Br + (b) In which solvent would this reaction proceed the fastest (Circle one) Methanol Acetone (c) Imagine that you are working for a chemical company and it was your job to perform a similar reaction to the one above, with the exception of the S atom in this reaction being replaced by an O atom. During the reaction, you observe the formation of three separate molecules instead of the single molecule obtained above. What is the likeliest other products that are formed? Draw them in the box provided.arrow_forward3. For the reactions below, draw the arrows corresponding to the transformations and draw in the boxes the reactants or products as indicated. Note: Part A should have arrows drawn going from the reactants to the middle structure and the arrows on the middle structure that would yield the final structure. For part B, you will need to draw in the reactant before being able to draw the arrows corresponding to product formation. A. B. Rearrangement ΘΗarrow_forward
- 2. Draw the arrows required to make the following reactions occur. Please ensure your arrows point from exactly where you want to exactly where you want. If it is unclear from where arrows start or where they end, only partial credit will be given. Note: You may need to draw in lone pairs before drawing the arrows. A. B. H-Br 人 C Θ CI H Cl Θ + Br Oarrow_forward4. For the reactions below, draw the expected product. Be sure to indicate relevant stereochemistry or formal charges in the product structure. a) CI, H e b) H lux ligh Br 'Harrow_forwardArrange the solutions in order of increasing acidity. (Note that K (HF) = 6.8 x 10 and K (NH3) = 1.8 × 10-5) Rank solutions from least acidity to greatest acidity. To rank items as equivalent, overlap them. ▸ View Available Hint(s) Least acidity NH&F NaBr NaOH NH,Br NaCIO Reset Greatest acidityarrow_forward
- 1. Consider the following molecular-level diagrams of a titration. O-HA molecule -Aion °° о ° (a) о (b) (c) (d) a. Which diagram best illustrates the microscopic representation for the EQUIVALENCE POINT in a titration of a weak acid (HA) with sodium. hydroxide? (e)arrow_forwardAnswers to the remaining 6 questions will be hand-drawn on paper and submitted as a single file upload below: Review of this week's reaction: H₂NCN (cyanamide) + CH3NHCH2COOH (sarcosine) + NaCl, NH4OH, H₂O ---> H₂NC(=NH)N(CH3)CH2COOH (creatine) Q7. Draw by hand the reaction of creatine synthesis listed above using line structures without showing the Cs and some of the Hs, but include the lone pairs of electrons wherever they apply. (4 pts) Q8. Considering the Zwitterion form of an amino acid, draw the Zwitterion form of Creatine. (2 pts) Q9. Explain with drawing why the C-N bond shown in creatine structure below can or cannot rotate. (3 pts) NH2(C=NH)-N(CH)CH2COOH This bond Q10. Draw two tautomers of creatine using line structures. (Note: this question is valid because problem Q9 is valid). (4 pts) Q11. Mechanism. After seeing and understanding the mechanism of creatine synthesis, students should be ready to understand the first half of one of the Grignard reactions presented in a past…arrow_forwardPropose a synthesis pathway for the following transformations. b) c) d)arrow_forward
- The rate coefficient of the gas-phase reaction 2 NO2 + O3 → N2O5 + O2 is 2.0x104 mol–1 dm3 s–1 at 300 K. Indicate whether the order of the reaction is 0, 1, or 2.arrow_forward8. Draw all the resonance forms for each of the following molecules or ions, and indicate the major contributor in each case, or if they are equivalent. (4.5 pts) (a) PH2 سمةarrow_forward3. Assign absolute configuration (Rors) to each chirality center. a. H Nitz C. он b. 0 H-C. C H 7 C. ་-4 917-417 refs H 1つ ८ ડુ d. Но f. -2- 01 Ho -OH 2HNarrow_forward
